I absolutely love tree problems!

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It had been a while since I had seen a fresh tree problem on Leetcode. This is a Medium one, but fairly straightforward - a direct implementation of a post-order depth-first-search traversal. Code is down below, cheers, ACC.

Count Nodes Equal to Sum of Descendants - LeetCode

1973. Count Nodes Equal to Sum of Descendants
Medium

Given the root of a binary tree, return the number of nodes where the value of the node is equal to the sum of the values of its descendants.

A descendant of a node x is any node that is on the path from node x to some leaf node. The sum is considered to be 0 if the node has no descendants.

 

Example 1:

Input: root = [10,3,4,2,1]
Output: 2
Explanation:
For the node with value 10: The sum of its descendants is 3+4+2+1 = 10.
For the node with value 3: The sum of its descendants is 2+1 = 3.

Example 2:

Input: root = [2,3,null,2,null]
Output: 0
Explanation:
No node has a value that is equal to the sum of its descendants.

Example 3:

Input: root = [0]
Output: 1
For the node with value 0: The sum of its descendants is 0 since it has no descendants.

 

Constraints:

  • The number of nodes in the tree is in the range [1, 105].
  • 0 <= Node.val <= 105

public int EqualToDescendants(TreeNode root)
{
    int sumDescendants = 0;
    int retVal = 0;

    EqualToDescendants(root, ref sumDescendants, ref retVal);
    return retVal;
}

private void EqualToDescendants(TreeNode currentNode, ref int sumDescendants, ref int retVal)
{
    if (currentNode == null)
    {
        sumDescendants = 0;
        return;
    }

    int sumDescendantsLeft = 0;
    EqualToDescendants(currentNode.left, ref sumDescendantsLeft, ref retVal);

    int sumDescendantsRight = 0;
    EqualToDescendants(currentNode.right, ref sumDescendantsRight, ref retVal);

    if (currentNode.val == sumDescendantsLeft + sumDescendantsRight)
    {
        retVal++;
    }

    sumDescendants = currentNode.val + sumDescendantsLeft + sumDescendantsRight;
}

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