112.HasPathSum.cs

// 112. Path Sum
// Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that
// adding up all the values along the path equals targetSum.
// A leaf is a node with no children.
// Example 1:
// Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
// Output: true
// Explanation: The root-to-leaf path with the target sum is shown.
// Example 2:
// Input: root = [1,2,3], targetSum = 5
// Output: false
// Explanation: There two root-to-leaf paths in the tree:
// (1 --> 2): The sum is 3.
// (1 --> 3): The sum is 4.
// There is no root-to-leaf path with sum = 5.
// Example 3:
// Input: root = [], targetSum = 0
// Output: false
// Explanation: Since the tree is empty, there are no root-to-leaf paths.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public bool HasPathSum(TreeNode root, int targetSum) {
        if(root == null)
            return false;
        if(root.left == null && root.right == null && targetSum == root.val)
            return true;
        if(root.left == null && root.right == null && targetSum != root.val)
            return false;
        return HasPathSum(root.left, targetSum-root.val) || 
            HasPathSum(root.right, targetSum-root.val);
    }
}