标签：etc 功能 void target bsp each public where solution
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 andsum = 22,
5
/ 4 8
/ / 11 13 4
/ \ / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
题意：给定一数，在树中找出所有路径和等于该数的情况。
方法一：
使用vector向量实现stack的功能，以方便输出指定路径。思想和代码和Path sum大致相同。
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int> > pathSum(TreeNode *root, int sum) { vector<vector<int>> res; vector<TreeNode *> vec; TreeNode *pre=NULL; TreeNode *cur=root; int temVal=0; while(cur|| !vec.empty()) { while(cur) { vec.push_back(cur); temVal+=cur->val; cur=cur->left; } cur=vec.back(); if(cur->left==NULL&&cur->right==NULL&&temVal==sum) { //和Path sum最大的区别 vector<int> temp; for(int i=0;i<vec.size();++i) temp.push_back(vec[i]->val); res.push_back(temp); } if(cur->right&&cur->right !=pre) cur=cur->right; else { vec.pop_back(); temVal-=cur->val; pre=cur; cur=NULL; } } return res; } };
方法二：递归法
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int> > pathSum(TreeNode *root, int sum) { vector<vector<int>> res; vector<int> path; findPaths(root,sum,res,path); return res; } void findPaths(TreeNode *root,int sum,vector<vector<int>> &res,vector<int> &path) { if(root==NULL) return; path.push_back(root->val); if(root->left==NULL&&root->right==NULL&&sum==root->val) res.push_back(path); findPaths(root->left,sum-root->val,res,path); findPaths(root->right,sum-root->val,res,path); path.pop_back(); } };
标签：etc 功能 void target bsp each public where solution
原文地址：http://www.cnblogs.com/love-yh/p/6980241.html
