第63天。
赶算法实验,再水一题。
今天的题目是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,
1 52 / \3 4 84 / / \5 11 13 46 / \ \7 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
比较简单,但是有一些坑点。
- 它要求一定要到
left
. - 然后空节点不能当成
0
.
然后是代码:
1bool hasPathSum(TreeNode* root, int sum) {2 if (!root ) return false;3 if (!root->left && !root->right) return sum == root->val;4 if (!root->left) return hasPathSum(root->right,sum-root->val);5 if (!root->right) return hasPathSum(root->left,sum - root->val);6 return hasPathSum(root->left,sum - root->val) || hasPathSum(root->right,sum - root->val);7}
其实dicuss
中的更精炼一点:
1bool hasPathSum(TreeNode *root, int sum) {2 if (root == NULL) return false;3 if (root->val == sum && root->left == NULL && root->right == NULL) return true;4 return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);5}