[leedcode 112] 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 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) {
        //空根节点默认false
        //注意递归终止条件：叶子节点，并且满足条件值（非叶子节点都不满足，还需要继续查询）
        //隐含条件root==null时 返回false！！
        if(root==null) return false;
        if(root.left==null&&root.right==null) return root.val==sum?true:false;
/*        else if(root.left==null) return hasPathSum(root.right,sum-root.val);
        else if(root.right==null) return hasPathSum(root.left,sum-root.val);*/
        else return hasPathSum(root.left,sum-root.val)||hasPathSum(root.right,sum-root.val);
    }
}