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Copy path110.平衡二叉树.go
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110.平衡二叉树.go
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/*
* @lc app=leetcode.cn id=110 lang=golang
*
* [110] 平衡二叉树
*
* https://leetcode-cn.com/problems/balanced-binary-tree/description/
*
* algorithms
* Easy (56.79%)
* Likes: 935
* Dislikes: 0
* Total Accepted: 326.6K
* Total Submissions: 574.9K
* Testcase Example: '[3,9,20,null,null,15,7]'
*
* 给定一个二叉树,判断它是否是高度平衡的二叉树。
*
* 本题中,一棵高度平衡二叉树定义为:
*
*
* 一个二叉树每个节点 的左右两个子树的高度差的绝对值不超过 1 。
*
*
*
*
* 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
* -10^4
*
*
*/
// @lc code=start
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func isBalanced(root *TreeNode) bool {
if root == nil {
return true
}
return isBalanced(root.Left) && isBalanced(root.Right) && abs(height(root.Left)-height(root.Right)) <= 1
}
func height(n *TreeNode) int {
if n == nil {
return 0
}
return max(height(n.Left), height(n.Right)) + 1
}
func max(x, y int) int {
if x > y {
return x
}
return y
}
func abs(x int) int {
if x < 0 {
return -1 * x
}
return x
}
// @lc code=end