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;
            else
                return 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;
    }
};