数据结构与算法之二叉树: LeetCode 110. 平衡二叉树 (Ts版)

平衡二叉树

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

描述

• 给定一个二叉树，判断它是否是 平衡二叉树

示例 1

输入：root = [3,9,20,null,null,15,7]
输出：true

示例 2

输入：root = [1,2,2,3,3,null,null,4,4]
输出：false

示例 3

输入：root = []
输出：true

提示

• 树中的节点数在范围 [0, 5000] 内
• - 1 0 4 10^4 104 <= Node.val <= 1 0 4 10^4 104

Typescript 版算法实现

1 ) 方案1: 自顶向下的递归

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function isBalanced(root: TreeNode | null): boolean {
    if (!root) return true;

    const height = (node: TreeNode | null): number => {
        if (!node) return 0;
        return Math.max(height(node.left), height(node.right)) + 1;
    }

    const leftHeight = height(root.left);
    const rightHeight = height(root.right);

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

2 ) 方案2: 自底向上的递归

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function isBalanced(root: TreeNode | null): boolean {
    function height(node: TreeNode | null): number {
        if (!node) return 0;
        const leftHeight = height(node.left);
        const rightHeight = height(node.right);
        // 如果左右子树中任意一个不平衡，或者高度差超过1，则当前树也不平衡
        if (leftHeight === -1 || rightHeight === -1 || Math.abs(leftHeight - rightHeight) > 1) {
            return -1;
        }
        return Math.max(leftHeight, rightHeight) + 1;
    }

    return height(root) !== -1;
}