Root Equals Sum of Children

Problem

You are given the root of a binary tree that consists of exactly 3 nodes: the root, its left child, and its right child.

Return true if the value of the root is equal to the sum of the values of its two children, or false otherwise.

Examples

Example 1:

Input: root = [10,4,6] Output: true Explanation: The values of the root, its left child, and its right child are 10, 4, and 6, respectively. 10 is equal to 4 + 6, so we return true.

Example 2:

Input: root = [5,3,1] Output: false Explanation: The values of the root, its left child, and its right child are 5, 3, and 1, respectively. 5 is not equal to 3 + 1, so we return false.

Constraints

• The tree consists only of the root, its left child, and its right child.
• -100 &lt= Node.val &lt= 100

Approach

1.Traversal: Since the tree structure is fixed with exactly three nodes, we can directly access the root and its children. 2.Comparison: Check if the value of the root node equals the sum of its left and right children.

Solution

Code in Different Languages

C++ Solution

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

class Solution {
public:
    bool isSumEqual(TreeNode* root) {
        if (!root) return false;
        int leftVal = (root->left) ? root->left->val : 0;
        int rightVal = (root->right) ? root->right->val : 0;
        return root->val == leftVal + rightVal;
    }
};

Java Solution

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;
    
    TreeNode(int val) {
        this.val = val;
    }
}

public class Solution {
    public boolean isSumEqual(TreeNode root) {
        if (root == null) return false;
        int leftVal = (root.left != null) ? root.left.val : 0;
        int rightVal = (root.right != null) ? root.right.val : 0;
        return root.val == leftVal + rightVal;
    }
}

Python Solution

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

class Solution:
    def isSumEqual(self, root: TreeNode) -> bool:
        if not root:
            return False
        left_val = root.left.val if root.left else 0
        right_val = root.right.val if root.right else 0
        return root.val == left_val + right_val

Complexity Analysis

Time Complexity: $O(1)$

Reason: for accessing the value of root node and its children , performing comparisons

Space Complexity: $O(1)$

Reason: we are not using any additional data structures.