LeetCode 1382. Balance a Binary Search Tree

原题链接在这里：https://leetcode.com/problems/balance-a-binary-search-tree/

题目：

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

题解：

Could perform a inorder traverse first and then construst balanced BST from inorder list.

TimeComplexity: O(n). n is total number of numbers in the tree.

Space: O(n). list size.

AC Java:

 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> list = new ArrayList<>();
18     public TreeNode balanceBST(TreeNode root) {
19         inorderDfs(root);
20         return constructBalancedBst(0, list.size() - 1);
21     }
22     
23     private void inorderDfs(TreeNode root){
24         if(root == null){
25             return;
26         }
27         
28         inorderDfs(root.left);
29         list.add(root);
30         inorderDfs(root.right);
31     }
32     
33     private TreeNode constructBalancedBst(int l, int r){
34         if(l > r){
35             return null;
36         }
37         
38         int mid = l + (r - l) / 2;
39         TreeNode root = list.get(mid);
40         root.left = constructBalancedBst(l, mid - 1);
41         root.right = constructBalancedBst(mid + 1, r);
42         return root;
43     }
44 }

类似Convert Sorted Array to Binary Search Tree.