leetcode_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.

只需要三行，當爲空的時候返回，等於返回true，不等於返回false。其他的沒到最後的時候，迭代，和減去當前值即可。代碼：

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