Leetcode 257. Binary Tree Paths

Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

 

   1
 /   \
2     3
 \
  5

 

All root-to-leaf paths are:

["1->2->5", "1->3"]
分析:典型的深度优先搜索,dfs函数思路为:
dfs函数中参数传入一个string,该String将每次结点的值连接起来,直到递归出口的时候返回;
当该结点有左孩子的时候,将左孩子的值连接到字符串尾部;
当该结点有右孩子的时候,将右孩子的值连接到字符串尾部;
当该结点无左右孩子的时候,将字符串push入数组后return;
 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     void dfs(vector<string> &v, TreeNode* root, string s){
13         if(root -> left == NULL && root -> right == NULL){
14             v.push_back(s);
15             return;
16         }
17         if(root -> left != NULL)
18             dfs(v, root -> left, s + "->" + to_string(root -> left -> val));
19         if(root -> right != NULL)
20             dfs(v, root -> right, s + "->" + to_string(root -> right -> val));
21     }
22     vector<string> binaryTreePaths(TreeNode* root) {
23         vector<string> v;
24         if(root == NULL)
25             return v;
26         dfs(v, root, to_string(root -> val));
27         return v;
28     }
29 };

 

 
posted @ 2016-08-28 17:10  琴影  阅读(735)  评论(0编辑  收藏  举报