437

Author: tech-cow

README.md

@@ -92,6 +92,9 @@ Success is like pregnancy, Everybody congratulates you but nobody knows how many
 |226|[Invert Binary Tree](https://leetcode.com/problems/invert-binary-tree/#/description)| [Python [Yu]](./tree/Yu/226.py) | _O(N)_| _O(1)_  | Easy | |[公瑾讲解](https://youtu.be/oiX3mqcAK0s)|
 |543|[Diameter of Binary Tree](https://leetcode.com/problems/diameter-of-binary-tree/#/description)| [Python [Yu]](./tree/Yu/543.py) | _O(N)_| _O(h)_  | Easy | |[公瑾讲解](https://www.youtube.com/watch?v=0VnOfu2pYTo)|
 |501|[Find Mode in Binary Search Tree](https://leetcode.com/problems/find-mode-in-binary-search-tree/#/description)| [Python [Yu]](./tree/Yu/501.py) | _O(N)_| _O(N)_  | Easy | |[公瑾讲解](https://youtu.be/v4F4x_uwMb8)|
+|572|[Subtree of Another Tree](https://leetcode.com/problems/subtree-of-another-tree/#/description)| [Python [Yu]](./tree/Yu/572.py) | _O(S*T)_| _O(1)_  | Easy | |[公瑾讲解](https://youtu.be/v4F4x_uwMb8)|
+|437|[Path Sum III](https://leetcode.com/problems/path-sum-iii/#/description)| [Python [Yu]](./tree/Yu/437.py) | _O(N^2)_| _O(1)_  | Easy | |[公瑾讲解](https://youtu.be/NTyOEYYyv-o)|
+
 
 |563|[Binary Tree Tilt](https://leetcode.com/problems/binary-tree-tilt/#/description)| [Python [Yu]](./tree/Yu/563.py) | _O(N)_| _O(1)_  | Easy | |[公瑾讲解](https://youtu.be/47FQVP4ynk0)|
 |257|[Binary Tree Paths](https://leetcode.com/problems/binary-tree-paths/#/description)| [Python [Yu]](./tree/Yu/257.py) | _O(N)_| _O(N)_  | Easy | |[公瑾讲解](https://youtu.be/Zr_7qq2f16k)|

tree/Yu/102.py

@@ -11,25 +11,24 @@
 
 
 class Solution(object):
-    #def __init__(self):
-     #   self.res = []
-
     def levelOrder(self, root):
         """
         :type root: TreeNode
         :rtype: List[List[int]]
         """
-        res = []
-        self.dfs_level(root, 0, res)
-        return res
+        self.res = []
+        def dfs(root, level):
+            # Edge
+            if not root:
+                return []
+            # Process
+            if level >= len(self.res):
+                self.res.append([])
+            self.res[level].append(root.val)
+
+            # Recursion
+            dfs(root.left, level + 1)
+            dfs(root.right, level + 1)
 
-    def dfs_level(self, node, level, res):
-        #Edge
-        if not node:
-            return
-        #res: [[3],[9, 20], [15,7 ]]
-        if level >= len(res):
-            res.append([])
-        res[level].append(node.val)
-        self.dfs_level(node.left, level+1, res)
-        self.dfs_level(node.right, level+1, res)
+        dfs(root, 0)
+        return self.res

tree/Yu/437.py

@@ -0,0 +1,40 @@
+#!/usr/bin/python
+# -*- coding: utf-8 -*-
+
+# Author: Yu Zhou
+# 437. Path Sum III
+
+# Definition for a binary tree node.
+# class TreeNode(object):
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution(object):
+    def pathSum(self, root, sum):
+        """
+        :type root: TreeNode
+        :type sum: int
+        :rtype: int
+        """
+
+        # Edge Case:
+        if not root:
+            return 0
+
+        # Process
+        def dfs(root, sum):
+            count = 0
+
+            if not root:
+                return 0
+            if root.val == sum:
+                count += 1
+
+            count += dfs(root.left, sum - root.val)
+            count += dfs(root.right, sum -root.val)
+            return count
+
+        # Recursion
+        return dfs(root,sum) + self.pathSum(root.left, sum) + self.pathSum(root.right, sum)

tree/Yu/572.py

@@ -0,0 +1,33 @@
+#!/usr/bin/python
+# -*- coding: utf-8 -*-
+
+# Author: Yu Zhou
+# 572. Subtree of Another Tree
+
+# ****************
+# Descrption:
+# ****************
+
+class Solution(object):
+    def isSubtree(self, s, t):
+        """
+        :type s: TreeNode
+        :type t: TreeNode
+        :rtype: bool
+        """
+        #Edge
+        if not s or not t:
+            return False
+
+        #Process
+        if self.isMatch(s,t):
+            return True
+
+        #Recusion
+        return self.isSubtree(s.left, t) or self.isSubtree(s.right, t)
+
+    def isMatch(self, s, t):
+        if not s or not t:
+            return s == t
+
+        return s.val == t.val and self.isMatch(s.left, t.left) and self.isMatch(s.right, t.right)