题目

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.

分析

题意：将二叉搜索树平衡；每个节点的两个子树深度不超过一个节点，称为平衡。

一个笨办法是先获取原树的所有值，然后根据二叉搜索树的规则(二分法)，重新建立起一颗平衡的二叉搜索树。

解答

class Solution {
    List<TreeNode> sortedArr = new ArrayList<>();
    public TreeNode balanceBST(TreeNode root) {
    	// 中序遍历得到的就是排好序的
        inorderTraverse(root);
        // 通过二分法重新构造二叉搜索树
        return sortedArrayToBST(0, sortedArr.size() - 1);
    }
    void inorderTraverse(TreeNode root) {
        if (root == null) return;
        inorderTraverse(root.left);
        sortedArr.add(root);
        inorderTraverse(root.right);
    }
    TreeNode sortedArrayToBST(int start, int end) {
        if (start > end) return null;
        int mid = (start + end) / 2;
        TreeNode root = sortedArr.get(mid);
        root.left = sortedArrayToBST(start, mid - 1);
        root.right = sortedArrayToBST(mid + 1, end);
        return root;
    }
}