112. Path Sum

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,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

Notice:

• Special case for empty tree and sum=0
• Four case none leaf node with only one child

/**
 * 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:
    bool fun(TreeNode* root, int sum){
        if(NULL == root){
            if(sum == 0) return true;
            else return false;
        }else{
            if(NULL == root->left && NULL == root->right){
                if(sum == root->val) return true;
                else return false;
            }else if(NULL == root->left && NULL != root->right){
                return fun(root->right, sum - root->val);
            }else if(NULL != root->left && NULL == root->right){
                return fun(root->left, sum - root->val);
            }else{
                return fun(root->left, sum - root->val) || fun(root->right, sum - root->val);
            }
        }
    }
    bool hasPathSum(TreeNode* root, int sum) {
        if(NULL == root) return false;
        else return fun(root, sum);
    }
};