Leetcoder - Path Sum

 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 exist a root-to-leaf path 5->4->11->2 which sum is 22.

/**
 * Definition for binary tree
 * 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 findPath(root, sum);
    }
 
    public boolean findPath(TreeNode node, int sum) {
        if(node.left==null && node.right==null) {
            if(node.val==sum) {
                return true;
            } else {
                return false;
            }
        } else if(node.left!=null && node.right!=null) {
            return findPath(node.left, sum - node.val) || findPath(node.right, sum - node.val);
        } else if(node.left==null && node.right!=null) {
            return findPath(node.right, sum - node.val);
        } else {
            return findPath(node.left, sum - node.val);
        }
    }
}


/**
 * 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 node, int sum) {
        if (node == null) {
            return false;
        }
        sum -= node.val;
        if (node.left == null && node.right == null && sum == 0) {
            return true;
        }
        return hasPathSum(node.left, sum) || hasPathSum(node.right, sum);
    }
}