Binary Tree Paths Solutions in C++

// Source : https://leetcode.com/problems/binary-tree-paths/
// Author : Calinescu Valentin, Hao Chen
// Date   : 2015-10-23


/**
 * 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:
    vector<string> TreePaths;
    void DFS(TreeNode* node, string answer)
    {
        answer += "->" + to_string(node->val);
        if(node->left == NULL && node->right == NULL)
            TreePaths.push_back(answer);
        else
        {
            if(node->left != NULL)
                DFS(node->left, answer);
            if(node->right != NULL)
                DFS(node->right, answer);
        }
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        if(root != NULL)
        {
            DFS(root, "");
            for(int i = 0; i < TreePaths.size(); i++)
                TreePaths[i].erase(TreePaths[i].begin(), TreePaths[i].begin() + 2);
        }
        return TreePaths;
    }
};




// Another more clear DFS implementation

class Solution {
public:
    void binaryTreePathsHelper(TreeNode* root, vector<int> solution, vector<string>& result ) {
        if (!root) return;
        
        solution.push_back(root->val);
        
        //meet the leaf node, shape a path into the result
        if (root->left==NULL && root->right==NULL){
            if(solution.size()>0){
                stringstream ss;
                for(int i=0; i<solution.size(); i++){
                    ss << solution[i] << (i<solution.size()-1 ? "->":"");
                }
                result.push_back(ss.str());
            }
            return;
        }
        
        binaryTreePathsHelper(root->left, solution, result);
        binaryTreePathsHelper(root->right, solution, result);
        
    }
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> result;
        vector<int> solution;
        binaryTreePathsHelper(root, solution, result);
        return result;
    }
};

/// Source : https://leetcode.com/problems/binary-tree-paths/description/
/// Author : liuyubobobo
/// Time   : 2017-11-17

#include <iostream>
#include <string>
#include <vector>

using namespace std;


/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


/// Recursive
/// Time Complexity: O(n), where n is the node's number in the tree
/// Space Complexity: O(h), where h is the height of the tree
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {

        vector<string> res;

        if(root == NULL)
            return res;

        if(root->left == NULL && root->right == NULL){
            res.push_back(to_string(root->val));
            return res;
        }

        vector<string> leftPaths = binaryTreePaths(root->left);
        for(int i = 0 ; i < leftPaths.size() ; i ++)
            res.push_back(to_string(root->val) + "->" + leftPaths[i]);

        vector<string> rightPaths = binaryTreePaths(root->right);
        for(int i = 0 ; i < rightPaths.size() ; i ++)
            res.push_back(to_string(root->val) + "->" + rightPaths[i]);

        return res;
    }
};


int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/binary-tree-paths/description/
/// Author : liuyubobobo
/// Time   : 2018-11-17

#include <iostream>
#include <string>
#include <vector>
#include <stack>

using namespace std;


/// Definition for a binary tree node.
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


/// Non-Recursive
/// Using Stack
///
/// Time Complexity: O(n), where n is the node's number in the tree
/// Space Complexity: O(h), where h is the height of the tree
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {

        vector<string> res;
        if(root == NULL)
            return res;

        stack<pair<TreeNode*, string>> stack;
        stack.push(make_pair(root, to_string(root->val)));
        while(!stack.empty()){
            TreeNode* cur = stack.top().first;
            string s = stack.top().second;
            stack.pop();

            if(!cur->left && !cur->right)
                res.push_back(s);

            if(cur->left)
                stack.push(make_pair(cur->left, s + "->" + to_string(cur->left->val)));
            if(cur->right)
                stack.push(make_pair(cur->right, s + "->" + to_string(cur->right->val)));
        }
        return res;
    }
};


int main() {

    return 0;
}