257. Binary Tree Paths

題目描述：

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

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

思路：深搜

程序：

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> v;
        string s;
        if(root == NULL) return v;
        else if(root->left == NULL && root->right == NULL){
            v.push_back(int2str(root->val));
            return v;
        }
        else{
            s = int2str(root->val);
            dfs(root, v, s);
        }
        return v;
    }
    string int2str(int n){
        string s;
        bool flag = false;
        if(n == 0) return "0";
        if(n < 0) {
            flag = true;
            n = -n;
        }
        while(n){
            s = (char)(n % 10 + '0') + s;
            n /= 10;
        }
        if(flag) s = "-" + s;
        return s;
    }
    void dfs(TreeNode* root, vector<string> &v, string &s){
        if(root->left == NULL && root->right == NULL){
            v.push_back(s);
            return;
        }
        if(root->left){
            string s1 = int2str(root->left->val);
            s = s + "->" + s1;
            dfs(root->left, v, s);
            s = s.substr(0, s.size()-s1.size()-2);
        }
        if(root->right){
            string s1 = int2str(root->right->val);
            s = s + "->" + s1;
            dfs(root->right, v, s);
            s = s.substr(0, s.size()-s1.size()-2);
        }
    }
};