【leetcode刷题笔记】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.

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题解：其实很类似这道题：http://www.cnblogs.com/sunshineatnoon/p/3853376.html，也是用递归的方法，在每个root上计算一个sum，表示从树根节点到当前节点得到的和，然后判断当前节点是否是叶节点，如果是，再判断从根节点到当前节点路径上的和是否等于sum，是就找到了要求的路径。

代码如下：

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     private boolean hadPath = false;
     private void hasPathDfs(TreeNode root,int sum,int currSum){
         if(root == null)
             return;
         currSum = currSum + root.val;
         if(root.left == null && root.right == null && currSum == sum){
             hadPath = true;
             return;
         }
         hasPathDfs(root.left, sum, currSum);
         hasPathDfs(root.right, sum, currSum);
     }
     public boolean hasPathSum(TreeNode root, int sum) {
         hasPathDfs(root, sum, 0);
         return hadPath;
     }
 }