剑指OFFER----面试题34. 二叉树中和为某一值的路径

链接:https://leetcode-cn.com/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof/

代码:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> res;
    vector<int> path;

    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        dfs(root, sum);
        return res;
    }

    void dfs(TreeNode* root, int sum) {
        if (!root) return;
        path.push_back(root->val);
        sum -= root->val;
        if (!root->left && !root->right && !sum) res.push_back(path);
        dfs(root->left, sum);
        dfs(root->right, sum);
        path.pop_back();  
    }

};

猜你喜欢

转载自www.cnblogs.com/clown9804/p/12386606.html