剑指 Offer 55 – II. 平衡二叉树
本问题对应的 leetcode 原文链接：剑指 Offer 55 – II. 平衡二叉树
问题描述
输入一棵二叉树的根节点，判断该树是不是平衡二叉树。如果某二叉树中任意节点的左右子树的深度相差不超过1，那么它就是一棵平衡二叉树。
示例 1:
给定二叉树 [3,9,20,null,null,15,7]
3
/ \
9 20
/ \
15 7
返回 true 。
示例 2:
给定二叉树 [1,2,2,3,3,null,null,4,4]
1
/ \
2 2
/ \
3 3
/ \
4 4
返回 false 。
限制：
0 <= 树的结点个数 <= 10000
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
// nlogn 空间 On 方法2:时间 On,空间O
public boolean isBalanced(TreeNode root) {
if(root == null) return true;
if(maxDepth(root) == -1) return false;
return true;
}
public int maxDepth(TreeNode root) {
if(root == null) return 0;
int left = maxDepth(root.left);
if(left == -1) return -1;
int right = maxDepth(root.right);
if(right == -1) return -1;
// 返回-1表示不符合条件了
if(Math.abs(left - right) > 1) return -1;
return Math.max(left, right) + 1;
}
}
Python
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def isBalanced(self, root):
"""
:type root: TreeNode
:rtype: bool
"""
if root is None:
return True
if self.maxDepth(root) == -1:
return False
return True
def maxDepth(self, root):
if root is None:
return 0
left = self.maxDepth(root.left)
if left == -1:
return -1
right = self.maxDepth(root.right)
if right == -1:
return -1
# 返回-1表示不符合条件了
if abs(left - right) > 1:
return -1
return max(left, right) + 1
C++
class Solution {
public:
bool isBalanced(TreeNode* root) {
if (root == nullptr) return true;
if (maxDepth(root) == -1) return false;
return true;
}
int maxDepth(TreeNode* root) {
if (root == nullptr) return 0;
int left = maxDepth(root->left);
if (left == -1) return -1;
int right = maxDepth(root->right);
if (right == -1) return -1;
if (abs(left - right) > 1) return -1;
return max(left, right) + 1;
}
};
Go
import "math"
func isBalanced(root *TreeNode) bool {
if root == nil {
return true
}
if maxDepth(root) == -1 {
return false
}
return true
}
func maxDepth(root *TreeNode) int {
if root == nil {
return 0
}
left := maxDepth(root.Left)
if left == -1 {
return -1
}
right := maxDepth(root.Right)
if right == -1 {
return -1
}
if math.Abs(float64(left-right)) > 1 {
return -1
}
return int(math.Max(float64(left), float64(right))) + 1
}
JS
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {boolean}
*/
var isBalanced = function(root) {
if (root == null) return true;
if (maxDepth(root) == -1) return false;
return true;
};
function maxDepth(root) {
if (root == null) return 0;
let left = maxDepth(root.left);
if (left == -1) return -1;
let right = maxDepth(root.right);
if (right == -1) return -1;
// 返回-1表示不符合条件了
if (Math.abs(left - right) > 1) return -1;
return Math.max(left, right) + 1;
}
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