# Univalue Tree Count - Amazon Top Interview Questions

### Problem Statement :

```A univalue tree is a tree where all nodes under it have the same value.

Given a binary tree root, return the number of univalue subtrees.

Constraints

1 ≤ n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [0, [1, null, null], [0, [1, [1, null, null], [1, null, null]], [0, null, null]]]

Output

5

Explanation

The unival trees include four leaf nodes (three of them are 1s, the other one is the rightmost 0), and one subtree in the middle (containing all 1s).

Example 2

Input

root = [1, [0, null, null], [0, null, null]]

Output

2

Explanation

The two leaf nodes are unival trees.```

### Solution :

```                        ```Solution in C++ :

bool dfs(Tree* node, int& cnt) {
if (node == NULL) {
return true;
}
bool ls = dfs(node->left, cnt);
bool rs = dfs(node->right, cnt);
int left_value = (node->left != NULL ? node->left->val : node->val);
int right_value = (node->right != NULL ? node->right->val : node->val);
if (node->val == left_value && node->val == right_value && ls && rs) {
cnt++;
return true;
}
return false;
}
int solve(Tree* root) {
int cnt = 0;
dfs(root, cnt);
return cnt;
}```
```

```                        ```Solution in Java :

import java.util.*;

/**
* public class Tree {
*   int val;
*   Tree left;
*   Tree right;
* }
*/
class Solution {
class Node {
int num;
boolean isUnivalued;
public Node(int n, boolean isUnivalued) {
this.num = n;
this.isUnivalued = isUnivalued;
}
}
int count = 0;

public int solve(Tree root) {
dfs(root);
return count;
}

private Node dfs(Tree node) {
if (node == null)
return null;
if (node.left == null && node.right == null) {
++count;
return new Node(node.val, true);
}

boolean isUnivalued = true;
if (node.left != null) {
Node left = dfs(node.left);
isUnivalued = (isUnivalued && left.isUnivalued && node.val == left.num);
}
if (node.right != null) {
Node right = dfs(node.right);
isUnivalued = (isUnivalued && right.isUnivalued && node.val == right.num);
}
if (isUnivalued)
++count;
return new Node(node.val, isUnivalued);
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, root):
self.cnt = 0

def rec(node):
if not node:
return True
left = not node.left or (node.left.val == node.val)
right = not node.right or (node.right.val == node.val)
left_sub = rec(node.left)
right_sub = rec(node.right)
if left and right and left_sub and right_sub:
self.cnt += 1
return True
return False

rec(root)
return self.cnt```
```

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