Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

      5
     / \
    4   8
   /   / \
  11  13  4
 /  \      \
7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public bool HasPathSum(TreeNode root, int sum) {
        //Edge case to detact, whenever root is null, return false
        if(root == null)
            return false;
        
        //Check if it is a leaf node, and it has val == root.val
        if(root.left == null && root.right == null)
        {
            if(root.val == sum)
                return true;
            else
                return false;
        }
        
        if(HasPathSum(root.left,sum-root.val))
        {
            return true;
        }
        if(HasPathSum(root.right,sum-root.val))
        {
            return true;
        }
        return false;
    }
}

Time Complexity : O(n)