README.md

965. 单值二叉树

English Version

题目描述

如果二叉树每个节点都具有相同的值，那么该二叉树就是单值二叉树。

只有给定的树是单值二叉树时，才返回 true；否则返回 false。

 

示例 1：

输入：[1,1,1,1,1,null,1]
输出：true

示例 2：

输入：[2,2,2,5,2]
输出：false

 

提示：

1. 给定树的节点数范围是 [1, 100]。
2. 每个节点的值都是整数，范围为 [0, 99] 。

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isUnivalTree(self, root: TreeNode) -> bool:
        def dfs(root):
            if root is None:
                return True
            if root.val != self.val:
                return False
            return dfs(root.left) and dfs(root.right)

        self.val = root.val
        return dfs(root)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private int val;

    public boolean isUnivalTree(TreeNode root) {
        val = root.val;
        return dfs(root);
    }

    private boolean dfs(TreeNode root) {
        if (root == null) {
            return true;
        }
        if (root.val != val) {
            return false;
        }
        return dfs(root.left) && dfs(root.right);
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int val;

    bool isUnivalTree(TreeNode* root) {
        val = root->val;
        return dfs(root);
    }

    bool dfs(TreeNode* root) {
        if (root == nullptr) return true;
        if (root->val != val) return false;
        return dfs(root->left) && dfs(root->right);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isUnivalTree(root *TreeNode) bool {
	return dfs(root, root.Val)
}

func dfs(root *TreeNode, val int) bool {
	if root == nil {
		return true
	}
	if root.Val != val {
		return false
	}
	return dfs(root.Left, val) && dfs(root.Right, val)
}

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