LeetCode：Path Sum II

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]
]

解題分析：

此題和Path Sum不太一樣，這道題是要把所有的路徑都求出來，而不只是返回一個找到與否的bool值。

要把路徑記錄下來，當一個結點的左右結點訪問完後返回時，應當把該結點剔除掉，也就是說存儲路徑的容器應當具備棧後進先出的特點，在c++中用vector也可以實現，另外還需要一個二維的向量來保存所有路徑。

代碼如下：

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