Path Sum II

题目名称
113. Path Sum II

描述
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]]

分析
  采用深度优先遍历，遍历根结点到叶子结点的每一条路径，将值等于目标值的路径保存。

C++代码

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<vector<int>> pathSum(TreeNode* root, int sum) {        vector<vector<int>> res;        vector<int> path;        DFS(root,sum,path,res);        return res;    }private:    void DFS(TreeNode *T,int sum,vector<int> &path,vector<vector<int>> &res) {       if(!T) return;       path.push_back(T->val);       if(T->val==sum && T->left==NULL &&T->right==NULL){           res.push_back(path);       }       DFS(T->left,sum-T->val,path,res);       DFS(T->right,sum-T->val,path,res);       path.pop_back();       return;    }};

总结
  最近在学习图的遍历算法，总算有点头绪了，这道题也是在参考了别人的答案之后自己写出来的。