# 965.Univalued Binary Tree

**965.Univalued Binary Tree**

难度:Easy

> 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： ![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/29/screen-shot-2018-12-25-at-50104-pm.png) 输入：\[1,1,1,1,1,null,1] 输出：true 示例 2：

![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2018/12/29/screen-shot-2018-12-25-at-50050-pm.png) 输入：\[2,2,2,5,2] 输出：false

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

代码如下：

```
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isUnivalTree(TreeNode* root) {
        if(!root) return true;
        if(!root->left && !root->right) return true;
        if(root->left && root->val!=root->left->val) return false;
        if(root->right && root->val !=root->right->val) return false;
        return isUnivalTree(root->left) && isUnivalTree(root->right);
        
    }
};
```

> 执行用时 : 16 ms, 在Univalued Binary Tree的C++提交中击败了4.04% 的用户 内存消耗 : 11.3 MB, 在Univalued Binary Tree的C++提交中击败了100.00% 的用户


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