Leetcode: 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.

 比较通俗易懂一点的看法：

/**
 * 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 helper(root, sum);
    }
    
    public boolean helper(TreeNode root, int remain) {
        if (root.left==null && root.right==null) {
            if (remain-root.val == 0) return true;
            else return false;
        }
        boolean left=false, right=false;
        if (root.left != null) {
            left = helper(root.left, remain-root.val);
        }
        if (root.right != null) {
            right = helper(root.right, remain-root.val);
        }
        return left || right;
    }
}

 精炼简洁的做法，但是不容易写：

public boolean hasPathSum(TreeNode root, int sum) {
    if(root == null)
        return false;
    if(root.left == null && root.right==null && root.val==sum)
        return true;
    return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);
}