Balance a Binary Search Tree

Given a binary search tree, return a balanced binary search tree with the same node values.

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

If there is more than one answer, return any of them.

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,null,null] is also correct.

Constraints:

• The number of nodes in the tree is between 1 and 10^4.
• The tree nodes will have distinct values between 1 and 10^5.

思路：首先用inorder 收集一个sorted list，然后build BST就可以了；

/**
 * 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 {
    public TreeNode balanceBST(TreeNode root) {
        List list = new ArrayList();
        inorder(root, list);
        return buildTree(list, 0, list.size() - 1);
    }
    
    private void inorder(TreeNode root, List list) {
        if(root == null) {
            return;
        }
        inorder(root.left, list);
        list.add(root.val);
        inorder(root.right, list);
    }
    
    private TreeNode buildTree(List list, int start, int end) {
        if(start > end) {
            return null;
        }
        if(start == end) {
            return new TreeNode(list.get(start));
        }
        int mid = start + (end - start) / 2;
        TreeNode root = new TreeNode(list.get(mid));
        root.left = buildTree(list, start, mid - 1);
        root.right = buildTree(list, mid + 1, end);
        return root;
    }
}