Path Sum | Leetcode

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.

For example:
Given the below binary tree and sum = 22,

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

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

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Solution: When the root node is null, return false (even when sum is 0). Recurse down the tree until you find a leaf node. For each recursion, subtract the value of the current node from sum. When we get to a leaf node, if the sum is 0, then return true.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) return false;
        return recurse(root, sum);
    }
    
    private boolean recurse(TreeNode node, int sum) {
        sum -= node.val;
        if (node.left == null && node.right == null) return (sum == 0);
        if (node.left != null && recurse(node.left, sum)) return true;
        if (node.right != null && recurse(node.right, sum)) return true;
        return false;
    }
}