LeetCode112 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. （Easy）

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.

分析：

还是用递归的思路，可以把sum - root -> val用在递归函数中，这样写最简洁。

代码：

 class Solution {

 public:

     bool hasPathSum(TreeNode* root, int sum) {

         if(root == nullptr) {

             return false;

         }

         if(root -> right == nullptr && root -> left == nullptr) {

             return sum == root->val;

         }

         return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val) ;

     }

 };

巴特西

LeetCode112 Path Sum

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