二叉樹--判斷二叉樹是否有和為sum的路徑

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.

Note: A leaf is a node with no children.

Example:

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

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