# [965. 单值二叉树](https://leetcode.cn/problems/univalued-binary-tree)

[English Version](/solution/0900-0999/0965.Univalued%20Binary%20Tree/README_EN.md)

## 题目描述

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<p>如果二叉树每个节点都具有相同的值，那么该二叉树就是<em>单值</em>二叉树。</p>

<p>只有给定的树是单值二叉树时，才返回&nbsp;<code>true</code>；否则返回 <code>false</code>。</p>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0965.Univalued%20Binary%20Tree/images/screen-shot-2018-12-25-at-50104-pm.png" style="height: 159px; width: 200px;"></p>

<pre><strong>输入：</strong>[1,1,1,1,1,null,1]
<strong>输出：</strong>true
</pre>

<p><strong>示例 2：</strong></p>

<p><img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0900-0999/0965.Univalued%20Binary%20Tree/images/screen-shot-2018-12-25-at-50050-pm.png" style="height: 158px; width: 200px;"></p>

<pre><strong>输入：</strong>[2,2,2,5,2]
<strong>输出：</strong>false
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ol>
	<li>给定树的节点数范围是&nbsp;<code>[1, 100]</code>。</li>
	<li>每个节点的值都是整数，范围为&nbsp;<code>[0, 99]</code>&nbsp;。</li>
</ol>

## 解法

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### **Python3**

<!-- 这里可写当前语言的特殊实现逻辑 -->

```python
# 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(node):
            if node is None:
                return True
            return node.val == root.val and dfs(node.left) and dfs(node.right)

        return dfs(root)
```

### **Java**

<!-- 这里可写当前语言的特殊实现逻辑 -->

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

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

### **C++**

```cpp
/**
 * 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:
    bool isUnivalTree(TreeNode* root) {
        return dfs(root, root->val);
    }

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

### **Go**

```go
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isUnivalTree(root *TreeNode) bool {
	var dfs func(*TreeNode) bool
	dfs = func(node *TreeNode) bool {
		if node == nil {
			return true
		}
		return node.Val == root.Val && dfs(node.Left) && dfs(node.Right)
	}
	return dfs(root)
}
```

### **TypeScript**

```ts
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function isUnivalTree(root: TreeNode | null): boolean {
    const val = root.val;
    const dfs = (root: TreeNode | null) => {
        if (root == null) {
            return true;
        }
        return root.val === val && dfs(root.left) && dfs(root.right);
    };
    return dfs(root.left) && dfs(root.right);
}
```

### **Rust**

```rust
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn dfs(val: i32, root: &Option<Rc<RefCell<TreeNode>>>) -> bool {
        if root.is_none() {
            return true;
        }
        let root = root.as_ref().unwrap().borrow();
        root.val == val && Self::dfs(val, &root.left) && Self::dfs(val, &root.right)
    }
    pub fn is_unival_tree(root: Option<Rc<RefCell<TreeNode>>>) -> bool {
        let root = root.as_ref().unwrap().borrow();
        Self::dfs(root.val, &root.left) && Self::dfs(root.val, &root.right)
    }
}
```

### **...**

```

```

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