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

该博客主要介绍了LeetCode上的965题，即判断一棵二叉树是否为单值二叉树。题目要求如果二叉树每个节点的值都相同，则返回true，否则返回false。博客提供了Python和Java两种解法，并提到Java解法的运行速度更快。

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

Example 2:
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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 = None

class Solution:
    def isUnivalTree(self, root: TreeNode) -> bool:
        if root==None:
            return True
        if (root.left!=None and root.left.val!=root.val) or (root.right!=None and root.right.val!=root.val):
            return False
        return 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.val
                and self.isUnivalTree(root.left))
        right_correct = (not root.right or root.val == root.right.val
                and self.isUnivalTree(root.right))
        return left_correct and right_correct