LeetCode_Path Sum II-优快云博客

本文链接：https://blog.youkuaiyun.com/sheng_ai/article/details/46695379

一.题目

Path Sum II

Total Accepted: 46778 Total Submissions: 175830 My Submissions

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]
]

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二.解题技巧

这道题和LeetCode_Path Sum类似，只不过这道题需要找到所有和等于给定值的路径，因此不能在中间部分进行剪枝，必须遍历完所有的路径。这里面有一个技巧，在进入每一个结点的时候，先将该结点的值push到vector中，在退出时间该结点的值pop出来，这样就可以避免有时会忘记pop结点的值的情况。

这种做法的时间复杂度为O(n),空间复杂度为O(logn)。

三.实现代码

#include <iostream>
#include <vector>

using std::vector;


/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/


struct TreeNode
{
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};


class Solution
{
private:
    void pathSum(TreeNode* root, int sum, vector<vector<int> > &Result,
                         vector<int> &TmpResult)
    {
        if (!root)
        {
            return;
        }

        if (!root->left && !root->right && root->val == sum)
        {
            TmpResult.push_back(sum);
            Result.push_back(TmpResult);

            // pop the leaf node
            TmpResult.pop_back();

            return;
        }

        int SumChild = sum - root->val;

        TmpResult.push_back(root->val);


        pathSum(root->left, SumChild, Result, TmpResult);
        pathSum(root->right, SumChild, Result, TmpResult);

        // pop the current node
        TmpResult.pop_back();
    }

public:
    vector<vector<int>> pathSum(TreeNode* root, int sum)
    {
        vector<vector<int> > Result;
        vector<int> TmpResult;

        pathSum(root, sum, Result, TmpResult);

        return Result;

    }
};