Path Sum

Problem:

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

Solution:


/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL)
            return false;
        if (root != NULL && root->left == NULL &&
            root->right == NULL && sum == root->val)
            return true;
        else {
            return hasPathSum(root->left, sum - root->val) ||
                hasPathSum(root->right, sum - root->val);
        }
    }
};

====== Second Try ========
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if (root == NULL)
            return false;
        else if (root -> val == sum && root -> left == NULL && root -> right == NULL)
            return true;
        else if (hasPathSum(root -> left, sum - root -> val) ||
                 hasPathSum(root -> right, sum - root -> val))
            return true;
        else
            return false;
    }
};

==== Clean one ====

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if (root == NULL)
            return false;
        else if (root -> val == sum && root -> left == NULL && root -> right == NULL)
            return true;
        else
            return hasPathSum(root -> left, sum - root -> val) ||
                 hasPathSum(root -> right, sum - root -> val);
    }
};