LeetCode257. Binary Tree Paths

文章目录

• • 一、题目
  • 二、题解

一、题目

Given the root of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.

Example 1:

Input: root = 1,2,3,null,5

Output: "1-\>2-\>5","1-\>3"

Example 2:

Input: root = 1

Output: "1"

Constraints:

The number of nodes in the tree is in the range 1, 100.

-100 <= Node.val <= 100

二、题解

前序遍历+回溯

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void getPath(TreeNode* root,vector<int>& path,vector<string>& res){
        path.push_back(root->val);
        if(root->left == nullptr && root->right == nullptr){
            string s;
            for(int i = 0;i < path.size();i++){
                s += to_string(path[i]);
                if(i != path.size() - 1) s += "->";
            }
            res.push_back(s);
        }
        if(root->left){
            getPath(root->left,path,res);
            path.pop_back();
        }
        if(root->right){
            getPath(root->right,path,res);
            path.pop_back();
        }
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<int> path;
        vector<string> res;
        getPath(root,path,res);
        return res;
    }
};