Sum of Left Leaves Solutions in C++

// Source : https://leetcode.com/problems/sum-of-left-leaves/
// Author : Hao Chen
// Date   : 2016-11-12



/**
 * 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 sumOfLeftLeaves_recursion_v1(TreeNode* root, int& result) {
        if (root == NULL ) {
            return;
        }
        
        if (root->left && root->left->left == NULL && root->left->right == NULL) {
            result += root->left->val;
        }
        sumOfLeftLeaves_recursion_v1(root->left, result);
        sumOfLeftLeaves_recursion_v1(root->right, result);
        
    }
    
    int sumOfLeftLeaves_recursion_v2(TreeNode* root) {
        if (root == NULL ) {
            return 0;
        }
        int result = 0;
        if (root->left && root->left->left == NULL && root->left->right == NULL) {
            result = root->left->val;
        }
        result += sumOfLeftLeaves_recursion_v2(root->left) + sumOfLeftLeaves_recursion_v2(root->right);
        return result;
    }    
    

    int sumOfLeftLeaves(TreeNode* root) {
        srand(time(NULL));
        if (rand()%2) {
            int result = 0;
            sumOfLeftLeaves_recursion_v1(root, result);
            return result;
        } else {
            return sumOfLeftLeaves_recursion_v2(root);
        }
        
    }
};

/// Source : https://leetcode.com/problems/sum-of-left-leaves/
/// Author : liuyubobobo
/// Time   : 2019-04-22

#include <iostream>

using namespace std;


/// Recursion
/// Time Complexity: O(n)
/// Space Complexity: O(h)

/// 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 {

private:
    int res = 0;

public:
    int sumOfLeftLeaves(TreeNode* root) {

        if(!root) return 0;

        dfs(root, false);
        return res;
    }

private:
    void dfs(TreeNode* node, bool isLeft){

        if(!node->left && !node->right){
            if(isLeft) res += node->val;
            return;
        }

        if(node->left) dfs(node->left, true);
        if(node->right) dfs(node->right, false);
    }
};


int main() {

    return 0;
}

/// Source : https://leetcode.com/problems/sum-of-left-leaves/
/// Author : liuyubobobo
/// Time   : 2019-04-22

#include <iostream>

using namespace std;


/// Recursion
/// Don't use class member variable
/// Time Complexity: O(n)
/// Space Complexity: O(h)

/// 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:
    int sumOfLeftLeaves(TreeNode* root) {

        if(!root) return 0;

        return dfs(root, false);
    }

private:
    int dfs(TreeNode* node, bool isLeft){

        if(!node->left && !node->right){
            if(isLeft) return node->val;
            return 0;
        }

        int res = 0;
        if(node->left) res += dfs(node->left, true);
        if(node->right) res += dfs(node->right, false);
        return res;
    }
};


int main() {

    return 0;
}