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Optional Binary Tree Morris Traversal Algorithms added.
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‎06-Stack-and-Queue/Course Code (C++)/Optional-04-Binary-Tree-Morris-Traversal/inorder.cpp

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/// Source : https://leetcode.com/problems/binary-tree-inorder-traversal/solution/
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/// Author : liuyubobobo
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/// Time : 2018-05-30
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#include <iostream>
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#include <vector>
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#include <stack>
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using namespace std;
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/// Definition for a binary tree node.
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struct TreeNode {
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int val;
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TreeNode *left;
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TreeNode *right;
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TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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};
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// InOrder Morris Traversal
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// Time Complexity: O(n), n is the node number in the tree
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// Space Complexity: O(1)
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class Solution {
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public:
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vector<int> inorderTraversal(TreeNode* root) {
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vector<int> res;
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if( root == NULL )
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return res;
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TreeNode* cur = root;
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while(cur != NULL){
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if(cur->left == NULL){
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res.push_back(cur->val);
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cur = cur->right;
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}
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else{
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TreeNode* prev = cur->left;
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while(prev->right != NULL && prev->right != cur)
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prev = prev->right;
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if(prev->right == NULL){
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prev->right = cur;
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cur = cur->left;
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}
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else{
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prev->right = NULL;
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res.push_back(cur->val);
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cur = cur->right;
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}
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}
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}
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return res;
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}
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};
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int main() {
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return 0;
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}

‎06-Stack-and-Queue/Course Code (C++)/Optional-04-Binary-Tree-Morris-Traversal/postorder.cpp

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/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
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/// Author : liuyubobobo
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/// Time : 2018-05-31
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#include <iostream>
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#include <vector>
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#include <stack>
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using namespace std;
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/// Definition for a binary tree node.
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struct TreeNode {
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int val;
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TreeNode *left;
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TreeNode *right;
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TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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};
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// Morris PostOrder Traversal
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//
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// Time Complexity: O(n)
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// Space Complexity: O(1)
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class Solution {
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public:
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vector<int> postorderTraversal(TreeNode* root) {
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vector<int> res;
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if(root == NULL)
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return res;
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TreeNode* dummyRoot = new TreeNode(-1);
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dummyRoot->left = root;
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TreeNode* cur = dummyRoot;
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while(cur != NULL){
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if(cur->left == NULL)
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cur = cur->right;
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else{
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TreeNode* prev = cur->left;
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while(prev->right != NULL && prev->right != cur)
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prev = prev->right;
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if(prev->right == NULL){
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prev->right = cur;
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cur = cur->left;
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}
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else{
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prev->right = NULL;
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reverseTraversal(cur->left, res);
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cur = cur->right;
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}
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}
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}
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delete dummyRoot;
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return res;
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}
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private:
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void reverseTraversal(TreeNode* node, vector<int>& res){
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int start = res.size();
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while(node != NULL){
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res.push_back(node->val);
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node = node->right;
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}
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reverse(res.begin() + start, res.end());
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}
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};
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void print_vec(const vector<int>& vec){
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for(int e: vec)
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cout << e << " ";
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cout << endl;
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}
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int main() {
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TreeNode* root = new TreeNode(1);
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root->right = new TreeNode(2);
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root->right->left = new TreeNode(3);
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print_vec(Solution().postorderTraversal(root));
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return 0;
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}
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‎06-Stack-and-Queue/Course Code (C++)/Optional-04-Binary-Tree-Morris-Traversal/preorder.cpp

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/// Source : https://leetcode.com/problems/binary-tree-preorder-traversal/description/
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/// Author : liuyubobobo
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/// Time : 2018-05-29
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#include <iostream>
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#include <vector>
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#include <stack>
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#include <cassert>
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using namespace std;
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/// Definition for a binary tree node.
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struct TreeNode {
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int val;
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TreeNode *left;
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TreeNode *right;
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TreeNode(int x) : val(x), left(NULL), right(NULL) {}
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};
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// PreOrder Morris Traversal
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// Time Complexity: O(n), n is the node number in the tree
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// Space Complexity: O(1)
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class Solution {
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public:
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vector<int> preorderTraversal(TreeNode* root) {
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vector<int> res;
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if(root == NULL)
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return res;
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TreeNode* cur = root;
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while(cur != NULL){
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if(cur->left == NULL){
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res.push_back(cur->val);
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cur = cur->right;
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}
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else{
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TreeNode* prev = cur->left;
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while(prev->right != NULL && prev->right != cur)
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prev = prev->right;
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if(prev->right == NULL){
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res.push_back(cur->val);
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prev->right = cur;
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cur = cur->left;
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}
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else{
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prev->right = NULL;
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cur = cur->right;
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}
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}
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}
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return res;
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}
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};
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int main() {
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return 0;
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}

‎06-Stack-and-Queue/Course Code (Java)/Optional-04-Binary-Tree-Morris-Traversal/src/InorderSolution.java

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/// Source : https://leetcode.com/problems/binary-tree-inorder-traversal/solution/
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/// Author : liuyubobobo
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/// Time : 2018-05-30
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import java.util.ArrayList;
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import java.util.List;
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import java.util.Stack;
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// Inorder Morris Traversal
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// Time Complexity: O(n), n is the node number in the tree
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// Space Complexity: O(1)
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public class InorderSolution {
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public List<Integer> inorderTraversal(TreeNode root) {
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ArrayList<Integer> res = new ArrayList<Integer>();
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if(root == null)
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return res;
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TreeNode cur = root;
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while(cur != null){
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if(cur.left == null){
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res.add(cur.val);
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cur = cur.right;
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}
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else{
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TreeNode prev = cur.left;
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while(prev.right != null && prev.right != cur)
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prev = prev.right;
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if(prev.right == null){
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prev.right = cur;
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cur = cur.left;
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}
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else{
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prev.right = null;
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res.add(cur.val);
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cur = cur.right;
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}
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}
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}
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return res;
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}
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}

‎06-Stack-and-Queue/Course Code (Java)/Optional-04-Binary-Tree-Morris-Traversal/src/PostorderSolution.java

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/// Source : https://leetcode.com/problems/binary-tree-postorder-traversal/description/
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/// Author : liuyubobobo
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/// Time : 2018-05-31
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import java.util.ArrayList;
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import java.util.Collections;
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import java.util.List;
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import java.util.Stack;
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// Morris PostOrder Traversal
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//
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// Time Complexity: O(n)
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// Space Complexity: O(1)
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public class PostorderSolution {
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public List<Integer> postorderTraversal(TreeNode root) {
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ArrayList<Integer> res = new ArrayList<Integer>();
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if(root == null)
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return res;
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TreeNode dummyRoot = new TreeNode(-1);
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dummyRoot.left = root;
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TreeNode cur = dummyRoot;
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while(cur != null){
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if(cur.left == null)
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cur = cur.right;
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else{
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TreeNode pre = cur.left;
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while(pre.right != null && pre.right != cur)
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pre = pre.right;
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if(pre.right == null){
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pre.right = cur;
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cur = cur.left;
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}
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else{
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pre.right = null;
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reverseTraversal(cur.left, res);
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cur = cur.right;
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}
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}
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}
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return res;
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}
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private void reverseTraversal(TreeNode node, ArrayList<Integer> res){
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int start = res.size();
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while(node != null){
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res.add(node.val);
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node = node.right;
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}
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int i = start, j = res.size() - 1;
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while(i < j){
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Integer t = res.get(i);
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res.set(i, res.get(j));
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res.set(j, t);
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i ++;
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j --;
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}
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}
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}

‎06-Stack-and-Queue/Course Code (Java)/Optional-04-Binary-Tree-Morris-Traversal/src/PreorderSolution.java

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/// Source : https://leetcode.com/problems/binary-tree-preorder-traversal/description/
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/// Author : liuyubobobo
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/// Time : 2018-05-29
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import java.util.ArrayList;
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import java.util.List;
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// PreOrder Morris Traversal
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// Time Complexity: O(n), n is the node number in the tree
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// Space Complexity: O(1)
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public class PreorderSolution {
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public List<Integer> preorderTraversal(TreeNode root) {
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ArrayList<Integer> res = new ArrayList<Integer>();
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if(root == null)
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return res;
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TreeNode cur = root;
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while(cur != null){
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if(cur.left == null){
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res.add(cur.val);
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cur = cur.right;
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}
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else{
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TreeNode prev = cur.left;
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while(prev.right != null && prev.right != cur)
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prev = prev.right;
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if(prev.right == null){
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res.add(cur.val);
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prev.right = cur;
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cur = cur.left;
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}
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else{
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prev.right = null;
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cur = cur.right;
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}
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}
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}
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return res;
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}
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}

‎06-Stack-and-Queue/Course Code (Java)/Optional-04-Binary-Tree-Morris-Traversal/src/TreeNode.java

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// Definition for a binary tree node.
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public class TreeNode {
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int val;
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TreeNode left;
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TreeNode right;
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TreeNode(int x) { val = x; }
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}

‎readme.md

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| 6-7 优先队列相关的算法问题 Top K Frequent Elements | [C++源码](https://github.com/liuyubobobo/Play-with-Algorithm-Interview/tree/master/06-Stack-and-Queue/Course%20Code%20(C%2B%2B)/07-Top-K-Frequent-Elements/) | [Java源码](06-Stack-and-Queue/Course%20Code%20(Java)/07-Top-K-Frequent-Elements/src/) |
9292
| 补充代码1：教科书经典前序非递归遍历 | [C++源码](06-Stack-and-Queue/Course%20Code%20(C%2B%2B)/Optional-01-Classic-Non-Recursive-Preorder-Traversal/) | [Java源码](06-Stack-and-Queue/Course%20Code%20(Java)/Optional-01-Classic-Non-Recursive-Preorder-Traversal/src/) |
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| 补充代码2：教科书经典中序非递归遍历 | [C++源码](06-Stack-and-Queue/Course%20Code%20(C%2B%2B)/Optional-02-Classic-Non-Recursive-Inorder-Traversal/) | [Java源码](06-Stack-and-Queue/Course%20Code%20(Java)/Optional-02-Classic-Non-Recursive-Inorder-Traversal/src/) |
94-
| 补充代码3：教科书经典后序非递归遍历 | [C++源码](06-Stack-and-Queue/Course%20Code%20(C%2B%2B)/Optional-03-Classic-Non-Recursive-Postorder-Traversal/) | [Java源码](06-Stack-and-Queue/Course%20Code%20(Java)/Optional-03-Classic-Non-Recursive-Postorder-Traversal/) |
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| 补充代码4：二叉树的Morris非递归遍历 | [C++源码] | [Java源码] |
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| 补充代码3：教科书经典后序非递归遍历 | [C++源码](06-Stack-and-Queue/Course%20Code%20(C%2B%2B)/Optional-03-Classic-Non-Recursive-Postorder-Traversal/) | [Java源码](06-Stack-and-Queue/Course%20Code%20(Java)/Optional-03-Classic-Non-Recursive-Postorder-Traversal/src/) |
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| 补充代码4：二叉树的Morris非递归遍历 | [C++源码](06-Stack-and-Queue/Course%20Code%20(C%2B%2B)/Optional-04-Binary-Tree-Morris-Traversal/) | [Java源码](06-Stack-and-Queue/Course%20Code%20(Java)/Optional-04-Binary-Tree-Morris-Traversal/src/) |
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| 补充代码5：双向广度优先搜索 Word Ladder | [C++源码] | [Java源码] |
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| **第七章：二叉树和递归** | [章节C++源码](https://github.com/liuyubobobo/Play-with-Algorithm-Interview/tree/master/07-Binary-Tree-and-Recursion/Course%20Code%20(C%2B%2B)/) | [章节Java源码](07-Binary-Tree-and-Recursion/Course%20Code%20(Java)/) |
9898
| 7-1 二叉树天然的递归结构 | [C++源码](https://github.com/liuyubobobo/Play-with-Algorithm-Interview/tree/master/07-Binary-Tree-and-Recursion/Course%20Code%20(C%2B%2B)/01-Maximum-Depth-of-Binary-Tree/) | [Java源码](07-Binary-Tree-and-Recursion/Course%20Code%20(Java)/01-Maximum-Depth-of-Binary-Tree/src) |

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