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[LeetCode] 1382. Balance a Binary Search Tree

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.

A binary search tree is balanced if the depth of the two subtrees of every node never differs by more than 1.

Example 1:

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.

Example 2:

Input: root = [2,1,3]
Output: [2,1,3]

Constraints:

The number of nodes in the tree is in the range [1, 104].
1 <= Node.val <= 105

将二叉搜索树变平衡。

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。

如果有多种构造方法，请你返回任意一种。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/balance-a-binary-search-tree
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这道题几乎和108题一样。108题已经给定了一个排好序的数组，这道题对数组没有排序。

既然我们处理的是一个二叉搜索树 BST，那么为了满足平衡的需求，我们首先需要将数组排序，把最中间的那个元素挑出来当做根节点，这样左右子树的节点个数才会平衡。然后我们再用类似108题的分治的做法，分别处理左右子树即可。

时间O(n)

空间O(n)

Java实现

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     List<TreeNode> sorted = new ArrayList<>();
18 
19     public TreeNode balanceBST(TreeNode root) {
20         inorder(root);
21         return helper(0, sorted.size() - 1);
22     }
23 
24     private void inorder(TreeNode root) {
25         if (root == null) {
26             return;
27         }
28         inorder(root.left);
29         sorted.add(root);
30         inorder(root.right);
31     }
32 
33     private TreeNode helper(int start, int end) {
34         if (start > end) {
35             return null;
36         }
37         int mid = start + (end - start) / 2;
38         TreeNode root = sorted.get(mid);
39         root.left = helper(start, mid - 1);
40         root.right = helper(mid + 1, end);
41         return root;
42     }
43 }