一.题目

Path Sum II

Total Accepted: 46778 Total Submissions: 175830My 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，仅仅只是这道题须要找到全部和等于给定值的路径，因此不能在中间部分进行剪枝。必须遍历全然部的路径。

这里面有一个技巧，在进入每个结点的时候。先将该结点的值push到vector中。在退出时间该结点的值pop出来，这样就能够避免有时会忘记pop结点的值的情况。

这样的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 nodeTmpResult.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 nodeTmpResult.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;}
};

四.体会

这道题和LeetCode_Path Sum类似，解法也是同样的吗。仅仅是不能进行剪枝而已。

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