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 :



title-img 




                        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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