Path Sum II (Find Path in Tree) -- LeetCode

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

 

思路：简单的DFS。要注意的一点是，所有的路径必须到达叶子结点才算数。叶子结点的定义是没有孩子的节点。假如root节点没有孩子节点，则root节点就算是叶子结点。但当root节点有一个左孩子节点却没有右孩子节点时，它就不能算是叶子节点了。

 class Solution {
 public:
     void help(vector<vector<int> >& res, TreeNode* root, vector<int>& path, int target)
     {
         TreeNode *left = root->left, *right = root->right;
         path.push_back(root->val);
         if (!left && !right && target == root->val)
             res.push_back(path);
         if (left)
             help(res, left, path, target - root->val);
         if (right)
             help(res, right, path, target - root->val);
         path.pop_back();
     }
     vector<vector<int>> pathSum(TreeNode* root, int sum) {
         vector<vector<int> > res;
         if (!root) return res;
         vector<int> path;
         help(res, root, path, sum);
         return res;
     }
 };