[Leetcode] 49. Path Sum II

题目

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For example:
Given the below binary tree and sum = 22,

          5
         / \
        4   8
       /   / \
      11  13  4
     /  \    / \
    7    2  5   1

return

[
[5,4,11,2],
[5,8,4,5]
]

解题之法

class Solution {
public:
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        vector<vector<int>> res;
        vector<int> out;
        helper(root, sum, out, res);
        return res;
    }
    void helper(TreeNode* node, int sum, vector<int>& out, vector<vector<int>>& res) {
        if (!node) return;
        out.push_back(node->val);
        if (sum == node->val && !node->left && !node->right) {
            res.push_back(out);
        }
        helper(node->left, sum - node->val, out, res);
        helper(node->right, sum - node->val, out, res);
        out.pop_back();
    }
};

分析

这道二叉树路径之和在之前的基础上又需要找出路径，但是基本思想都一样，还是需要用深度优先搜索DFS，只不过数据结构相对复杂一点，需要用到二维的vector，而且每当DFS搜索到新节点时，都要保存该节点。而且每当找出一条路径之后，都将这个保存为一维vector的路径保存到最终结果二位vector中。并且，每当DFS搜索到子节点，发现不是路径和时，返回上一个结点时，需要把该节点从一维vector中移除。