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;

/**
 * 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:
    void dfs(vector<string> &v, TreeNode* root, string s){
        if(root -> left == NULL && root -> right == NULL){
            v.push_back(s);
            return;
        }
        if(root -> left != NULL)
            dfs(v, root -> left, s + "->" + to_string(root -> left -> val));
        if(root -> right != NULL)
            dfs(v, root -> right, s + "->" + to_string(root -> right -> val));
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> v;
        if(root == NULL)
            return v;
        dfs(v, root, to_string(root -> val));
        return v;
    }
};