文章目錄

• 題目描述
• 思路 && 代碼
• • • 二刷

一臉懵逼，居然沒寫這道題的題解。。

題目描述

思路 && 代碼

• 思路：平衡二叉樹判斷公式 = 左子樹滿足 + 右子樹滿足 + 左右子樹高度差不超過1

/*** Definition for a binary tree node.* public class TreeNode {* int val;* TreeNode left;* TreeNode right;* TreeNode(int x) { val = x; }* }*/ class Solution {public boolean isBalanced(TreeNode root) {return isBalancedTree(root) != -1;}// 為平衡二叉樹，返回深度；否則返回-1int isBalancedTree(TreeNode root){// 空為平衡二叉樹if(root == null){return 0;}int left = isBalancedTree(root.left);int right = isBalancedTree(root.right);// 左子樹不是 ｜｜ 右子樹不是 ｜｜ 左右子樹不平衡if(left == -1 || right == -1 || left - right < -1 || left - right > 1){return -1;}return Math.max(left, right) + 1;} }

• 無注釋版本

class Solution {public boolean isBalanced(TreeNode root) {return isBalancedTree(root) != -1; }public int isBalancedTree(TreeNode root) {if(root == null) {return 0;}int left = isBalancedTree(root.left);int right = isBalancedTree(root.right);if(left - right > 1 || left - right < -1 || left == -1 || right == -1) {return -1;}return Math.max(left, right) + 1;} }

二刷

• 好家伙，寫出了不一樣的寫法

class Solution {public boolean isBalanced(TreeNode root) {if(root == null) {return true;}if(isBalanced(root.left) && isBalanced(root.right)) {int left = isBalancedTree(root.left);int right = isBalancedTree(root.right);int diff = left - right;return diff < 2 && diff > -2;}return false;}int isBalancedTree(TreeNode root) {if(root == null) {return 0;}return Math.max(isBalancedTree(root.left), isBalancedTree(root.right)) + 1;} }

• 還是這個簡潔點哇

class Solution {public boolean isBalanced(TreeNode root) {return isBalancedTree(root) != -1;}int isBalancedTree(TreeNode root) {if(root == null) {return 0;}int left = isBalancedTree(root.left);int right = isBalancedTree(root.right);if(right - left > 1 || right - left < -1 || left == -1 || right == -1) {return -1;}return Math.max(left, right) + 1;} }

總結

以上是生活随笔為你收集整理的【LeetCode笔记】剑指 Offer 55 - II. 平衡二叉树（递归、二叉树）的全部內容，希望文章能夠幫你解決所遇到的問題。

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