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:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    List<TreeNode> list = new ArrayList<>();
    public TreeNode balanceBST(TreeNode root) {
        inorderDfs(root);
        return constructBalancedBst(0, list.size() - 1);
    }
    
    private void inorderDfs(TreeNode root){
        if(root == null){
            return;
        }
        
        inorderDfs(root.left);
        list.add(root);
        inorderDfs(root.right);
    }
    
    private TreeNode constructBalancedBst(int l, int r){
        if(l > r){
            return null;
        }
        
        int mid = l + (r - l) / 2;
        TreeNode root = list.get(mid);
        root.left = constructBalancedBst(l, mid - 1);
        root.right = constructBalancedBst(mid + 1, r);
        return root;
    }
}

类似Convert Sorted Array to Binary Search Tree.