We use cookies (including Google cookies) to personalize ads and analyze traffic. By continuing to use our site, you accept our Privacy Policy.

SimplyLeet
TagsDisclaimersPrivacyTermsContactAbout Us

Root Equals Sum of Children

Difficulty: Easy

Problem Description

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.

Key Insights

  • The problem involves a binary tree with exactly three nodes: one root node and two children (left and right).
  • The solution requires checking if the value of the root node is equal to the sum of its left and right children.
  • The constraints ensure that the input is always valid and simple, making it straightforward to implement.

Space and Time Complexity

Time Complexity: O(1) - The solution involves a constant number of operations regardless of the input size. Space Complexity: O(1) - No additional space is used apart from a few variables.

Solution

To solve this problem, we need to access the value of the root node and the values of its left and right children. The algorithm performs the following steps:

  1. Retrieve the value of the root node.
  2. Retrieve the values of the left and right children.
  3. Check if the root's value is equal to the sum of the left and right children's values.
  4. Return true if they are equal; otherwise, return false.

The data structure involved is a binary tree, and the algorithmic approach is simple arithmetic comparison.

Code Solutions

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

def checkTree(root):
    # Check if the root value equals the sum of its children's values
    return root.val == (root.left.val + root.right.val)
← Back to All Questions