题解 | 二叉树中和为某一值的路径(一)

/**
 * struct TreeNode {
 *	int val;
 *	struct TreeNode *left;
 *	struct TreeNode *right;
 *	TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (root == nullptr) return false;
        bool flag_res = false;
        function<void(TreeNode*, int)> dfs = [&](TreeNode* node, int cur_sum){
            if (node->left == nullptr && node->right == nullptr && (cur_sum+node->val) == sum) {
                flag_res = true;
                return;
            }
            cur_sum += node->val;   
            if (node->left) dfs(node->left, cur_sum);
            if (node->right) dfs(node->right, cur_sum);
        };
        dfs(root, 0);
        return flag_res;
    }
};