题目：
112. Path Sum (Easy)
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.
Note: A leaf is a node with no children.
Example:1
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10Given 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.
题目大意：
判断一个二叉树上，有没有从根root到叶leaf的路径上有没有结点值的和为给定数值的情况，并返回结果。
解题思路：
常规递归的写法，注意 “||”运算符：若运算符左边为true，则不再对运算符右侧进行运算。
代码：1
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9Java:
public boolean hasPathSum(TreeNode root,int sum){
if(root == null) return false;
if(root.left == null && root.right == null && root.val == sum) return true;
//注意这里的写法，因为需要从根到叶节点，所以要判断当前是否是叶节点
return hasPathSum(root.left,sum - root.val) || hasPathSum(root.right, sum - root.val);
}
