一刷1382

Author: diguage

README.adoc

@@ -9700,12 +9700,12 @@ TIP: **公众号的微信号是: `jikerizhi`**。__因为众所周知的原因
 //|Medium
 //|
 
-//|{counter:codes}
-//|{leetcode_base_url}/balance-a-binary-search-tree/[1382. Balance a Binary Search Tree^]
-//|{source_base_url}/_1382_BalanceABinarySearchTree.java[Java]
-//|{doc_base_url}/1382-balance-a-binary-search-tree.adoc[题解]
-//|Medium
-//|
+|{counter:codes}
+|{leetcode_base_url}/balance-a-binary-search-tree/[1382. Balance a Binary Search Tree^]
+|{source_base_url}/_1382_BalanceABinarySearchTree.java[Java]
+|{doc_base_url}/1382-balance-a-binary-search-tree.adoc[题解]
+|Medium
+|
 
 //|{counter:codes}
 //|{leetcode_base_url}/maximum-performance-of-a-team/[1383. Maximum Performance of a Team^]

docs/1382-balance-a-binary-search-tree.adoc

@@ -1,42 +1,42 @@
 [#1382-balance-a-binary-search-tree]
-= 1382. Balance a Binary Search Tree
+= 1382. 将二叉搜索树变平衡
 
-{leetcode}/problems/balance-a-binary-search-tree/[LeetCode - 1382. Balance a Binary Search Tree ^]
+https://leetcode.cn/problems/balance-a-binary-search-tree/[LeetCode - 1382. 将二叉搜索树变平衡 ^]
 
-Given the `root` of a binary search tree, return _a *balanced* binary search tree with the same node values_. If there is more than one answer, return *any of them*.
+给你一棵二叉搜索树，请你返回一棵 *平衡后* 的二叉搜索树，新生成的树应该与原来的树有着相同的节点值。如果有多种构造方法，请你返回任意一种。
 
-A binary search tree is *balanced* if the depth of the two subtrees of every node never differs by more than `1`.
+如果一棵二叉搜索树中，每个节点的两棵子树高度差不超过 `1`，我们就称这棵二叉搜索树是 *平衡的* 。
 
- 
-*Example 1:*
-<img alt="" src="https://assets.leetcode.com/uploads/2021/08/10/balance1-tree.jpg" style="width: 500px; height: 319px;" />
-[subs="verbatim,quotes"]
-----
-*Input:* root = [1,null,2,null,3,null,4,null,null]
-*Output:* [2,1,3,null,null,null,4]
-*Explanation:* This is not the only correct answer, [3,1,4,null,2] is also correct.
-----
+*示例 1：*
 
-*Example 2:*
-<img alt="" src="https://assets.leetcode.com/uploads/2021/08/10/balanced2-tree.jpg" style="width: 224px; height: 145px;" />
-[subs="verbatim,quotes"]
-----
-*Input:* root = [2,1,3]
-*Output:* [2,1,3]
-----
+image::images/1382-01.jpg[{image_attr}]
 
- 
-*Constraints:*
+....
+输入：root = [1,null,2,null,3,null,4,null,null]
+输出：[2,1,3,null,null,null,4]
+解释：这不是唯一的正确答案，[3,1,4,null,2,null,null] 也是一个可行的构造方案。
+....
 
+*示例 2：*
 
-* The number of nodes in the tree is in the range `[1, 10^4^]`.
-* `1 <= Node.val <= 10^5^`
+image::images/1382-02.jpg[{image_attr}]
 
+....
+输入: root = [2,1,3]
+输出: [2,1,3]
+....
+
+
+*提示：*
+
+* 树节点的数目在 `[1, 10^4^]` 范围内。
+* `1 \<= Node.val \<= 10^5^`
 
 
 
 == 思路分析
 
+二叉搜索树，先中序遍历拿到所有节点，然后递归构造平衡二叉搜索树。
 
 [[src-1382]]
 [tabs]
@@ -63,4 +63,6 @@ include::{sourcedir}/_1382_BalanceABinarySearchTree.java[tag=answer]
 
 == 参考资料
 
-
+. https://leetcode.cn/problems/balance-a-binary-search-tree/solutions/150820/shou-si-avlshu-wo-bu-guan-wo-jiu-shi-yao-xuan-zhua/[1382. 将二叉搜索树变平衡 - 手撕AVL树，我不管，我就是要旋转^]
+. https://leetcode.cn/problems/balance-a-binary-search-tree/solutions/457827/1382-jiang-er-cha-sou-suo-shu-bian-ping-heng-gou-z/[1382. 将二叉搜索树变平衡 - 「代码随想录」1382. 将二叉搜索树变平衡:【构造平衡二叉搜索树】详解^]
+. https://leetcode.cn/problems/balance-a-binary-search-tree/solutions/241897/jiang-er-cha-sou-suo-shu-bian-ping-heng-by-leetcod/[1382. 将二叉搜索树变平衡 - 官方题解^]

docs/images/1382-01.jpg

@@ -2847,7 +2847,7 @@ include::1362-closest-divisors.adoc[leveloffset=+1]
 
 // include::1381-design-a-stack-with-increment-operation.adoc[leveloffset=+1]
 
-// include::1382-balance-a-binary-search-tree.adoc[leveloffset=+1]
+include::1382-balance-a-binary-search-tree.adoc[leveloffset=+1]
 
 // include::1383-maximum-performance-of-a-team.adoc[leveloffset=+1]
 

docs/images/1382-02.jpg

@@ -883,6 +883,11 @@ endif::[]
 |{doc_base_url}/1023-camelcase-matching.adoc[题解]
 |✅ 双指针
 
+|{counter:codes2503}
+|{leetcode_base_url}/balance-a-binary-search-tree/[1382. 将二叉搜索树变平衡^]
+|{doc_base_url}/1382-balance-a-binary-search-tree.adoc[题解]
+|✅ 二叉搜索树，先中序遍历拿到所有节点，然后递归构造平衡二叉搜索树。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

docs/index.adoc

@@ -0,0 +1,55 @@
+package com.diguage.algo.leetcode;
+
+import com.diguage.algo.util.TreeNode;
+import com.diguage.util.TreeNodes;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+import java.util.List;
+
+public class _1382_BalanceABinarySearchTree {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-05-31 22:12:20
+   */
+  public TreeNode balanceBST(TreeNode root) {
+    List<TreeNode> data = new ArrayList<>();
+    TreeNode cur = root;
+    while (cur != null) {
+      TreeNode mostRight = cur.left;
+      if (mostRight != null) {
+        while (mostRight.right != null && mostRight.right != cur) {
+          mostRight = mostRight.right;
+        }
+        if (mostRight.right == null) {
+          mostRight.right = cur;
+          cur = cur.left;
+          continue;
+        } else {
+          mostRight.right = null;
+        }
+      }
+      data.add(cur);
+      cur = cur.right;
+    }
+    TreeNode result = dfs(data, 0, data.size() - 1);
+    return result;
+  }
+
+  private TreeNode dfs(List<TreeNode> data, int start, int end) {
+    if (start > end) {
+      return null;
+    }
+    int mid = start + (end - start) / 2;
+    TreeNode node = data.get(mid);
+    node.left = dfs(data, start, mid - 1);
+    node.right = dfs(data, mid + 1, end);
+    return node;
+  }
+  // end::answer[]
+  public static void main(String[] args) {
+    new _1382_BalanceABinarySearchTree()
+      .balanceBST(TreeNodes.buildTree(Arrays.asList(4, 2, 6, 1, 3, 5, 7)));
+  }
+}