# 965. Univalued Binary Tree

## Description

A binary tree is uni-valued if every node in the tree has the same value.

Given the root of a binary tree, return true if the given tree is uni-valued, or false otherwise.

Example 1:

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


Example 2:

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


Constraints:

• The number of nodes in the tree is in the range [1, 100].
• 0 <= Node.val < 100

## Solutions

• /**
* 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);
}
}

• /**
* 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);
}
};

• # 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)


• /**
* 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)
}

• /**
* 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);
}


• // 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)
}
}