題目描述

英文版描述

Given a binary tree, determine if it is height-balanced. For this problem, a height-balanced binary tree is defined as: a binary tree in which the left and right subtrees of every node differ in height by no more than 1.

英文版地址

https://leetcode.com/problems/balanced-binary-tree/

中文版描述

給定一個二叉樹，判斷它是否是高度平衡的二叉樹。 本題中，一棵高度平衡二叉樹定義為： 一個二叉樹每個節點 的左右兩個子樹的高度差的絕對值不超過 1 。

示例 1： 輸入：root = [3,9,20,null,null,15,7] 輸出：true

示例 2： 輸入：root = [1,2,2,3,3,null,null,4,4] 輸出：false

示例 3： 輸入：root = [] 輸出：true

提示：

• 樹中的節點數在範圍 [0, 5000] 內

• -104 <= Node.val <= 104

中文版地址

https://leetcode.cn/problems/balanced-binary-tree/

解題思路

利用二叉搜索樹的一個重要特性特性，即的左右子樹也分別為二叉搜索樹

解題方法

俺這版

/**
 * 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) {
        if (null == root) {
            return true;
        }
        int step = getHeight(root.left) - getHeight(root.right);

        return Math.abs(step) <= 1 && isBalanced(root.left) && isBalanced(root.right);
    }

    public int getHeight(TreeNode treeNode) {
        if (null == treeNode) {
            return 0;
        }
        return Math.max(getHeight(treeNode.left), getHeight(treeNode.right)) + 1;
    }
}

官方版

方法一：自頂向下的遞歸

class Solution {
    public boolean isBalanced(TreeNode root) {
        if (root == null) {
            return true;
        } else {
            return Math.abs(height(root.left) - height(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);
        }
    }

    public int height(TreeNode root) {
        if (root == null) {
            return 0;
        } else {
            return Math.max(height(root.left), height(root.right)) + 1;
        }
    }
}

方法二：自底向上的遞歸

class Solution {
    public boolean isBalanced(TreeNode root) {
        return height(root) >= 0;
    }

    public int height(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int leftHeight = height(root.left);
        int rightHeight = height(root.right);
        if (leftHeight == -1 || rightHeight == -1 || Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        } else {
            return Math.max(leftHeight, rightHeight) + 1;
        }
    }
}

總結

數據結構其實就是利用特殊的結構來增加操作（查找等）數據的效率o(^▽^)o