1382. 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.

class Solution {
    List<TreeNode> list = new ArrayList();
    public TreeNode balanceBST(TreeNode root) {
        inorder(root);
        return help(list, 0, list.size() - 1);
    }
    public void inorder(TreeNode root){
        if(root == null) return;
        inorder(root.left);
        list.add(root);
        inorder(root.right);
    }
    public TreeNode help(List<TreeNode> list, int start, int end){
        if(end < start) return null;
        int rootind = (end + start) / 2;
        TreeNode root = list.get(rootind);
        root.left = help(list, start, rootind - 1);
        root.right = help(list, rootind + 1, end);
        return root;
    }
}

把原来的bst按inorder存成sorted list，然后参考 https://leetcode.com/problems/convert-sorted-array-to-binary-search-tree/