LeetCode 965 Univalued Binary Tree--判断二叉树的所有节点的值是否相同--python,java解法

题目地址：Univalued Binary Tree - LeetCode

Acceptance: 67.6%
Difficulty: Easy

A binary tree is univalued if every node in the tree has the same value.

Return true if and only if the given tree is univalued.

Example 1:

Input: [1,1,1,1,1,null,1]
Output: true

Example 2:

Input: [2,2,2,5,2]
Output: false

Note:
The number of nodes in the given tree will be in the range [1, 100].
Each node’s value will be an integer in the range [0, 99].

这题的意思是判断一个二叉树的所有节点的值是否相同。

python3代码：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def isUnivalTree(self, root: TreeNode) -> bool:if root==None:return Trueif (root.left!=None and root.left.val!=root.val) or (root.right!=None and root.right.val!=root.val):return Falsereturn self.isUnivalTree(root.left) and self.isUnivalTree(root.right)

Java代码：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public boolean isUnivalTree(TreeNode root) {if (root==null){return true;}if( (root.left!=null && root.left.val!=root.val) || (root.right!=null && root.right.val!=root.val)){return false;}return isUnivalTree(root.left) && isUnivalTree(root.right);}
}

java运行就是比Python快，只要2ms。

官方参考解法：

class Solution(object):def isUnivalTree(self, root):left_correct = (not root.left or root.val == root.left.valand self.isUnivalTree(root.left))right_correct = (not root.right or root.val == root.right.valand self.isUnivalTree(root.right))return left_correct and right_correct

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