代码随想录：路径总和

代码随想录：路径总和

思路很简单，但是一个传参很巧妙，每次递归减去当前节点的值，就只用一个传参

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left),* right(right) {}* };*/
class Solution {
public:bool hasPathSum(TreeNode* root, int targetSum) {if (!root)return false;return tra(root, targetSum);}bool tra(TreeNode* node, int targetSum) {targetSum = targetSum - node->val;if (node->left == NULL && node->right == NULL) {if (targetSum == 0)return true;elsereturn false;}bool l = false, r = false;if (node->left) {l = tra(node->left, targetSum);}if (node->right) {r = tra(node->right, targetSum);}return l || r;}
};