🌗 110. 平衡二叉树

难度: 🌗

问题描述

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解法

/**
 * 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 boolean isBalanced(TreeNode root) {
        // 思路：
        // 递归 - 求当前节点左右子树的高度差，若在范围内，更新最大高度
        return mySol(root) >= 0;
    }

    private int mySol(TreeNode root) {
        // 递归终止条件
        if(root == null) {
            return 0;
        }
        // root != null
        int left = mySol(root.left);
        if(left < 0) {
            return -1;
        }
        int right = mySol(root.right);
        if(right < 0) {
            return -1;
        }
        if(Math.abs(left - right) > 1) {
            return -1;
        }
        return Math.max(left, right) + 1;
    }
}

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输出