257_Binary_Tree_Paths.cpp

/*
257. Binary Tree Paths
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
*/
/**
 * 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 {
private:
    void solve(TreeNode* root,vector<string> &ans,string str){
        if(!root) return;
        if(root && !root->left && !root->right){
            str+=to_string(root->val);
            ans.push_back(str);
            return;
        }
        str+=(to_string(root->val)+"->");
        solve(root->left,ans,str);
        solve(root->right,ans,str);
    }
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> ans;
        solve(root,ans,"");
        return ans;
    }
};