【leetcode】Path Sum II
2014-05-19 20:30
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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
return
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] ]
下午做了个笔试没睡觉,晚上整个人都不好了,一点刷题的感觉都没有。 很容易想到用深度有限搜索。开始想用栈实现,结果写的乱七八槽,后来才改成用递归实现深搜。 用数组path记录从根节点到当前的路径,如果当前节点是叶节点并且找到合适的路径,就把path转成vector放入结果的二维vector中;如果当前不是叶节点,就假定它在路径上,把它放入path中,并且把sum减掉当前节点的val,供递归时候使用。 代码如下:
1 #include <iostream> 2 #include <vector> 3 #include <stack> 4 using namespace std; 5 6 struct TreeNode { 7 int val; 8 TreeNode *left; 9 TreeNode *right; 10 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 11 }; 12 13 class Solution { 14 private: 15 int path[10000]; 16 vector<vector<int> > answer; 17 public: 18 vector<vector<int> > pathSum(TreeNode *root, int sum) { 19 dfs(root,sum,0); 20 return answer; 21 } 22 void dfs(TreeNode* root,int sum,int path_index){ 23 if(root == NULL) 24 return; 25 if(root->val == sum && root->left == NULL && root->right == NULL) 26 { 27 //找到一条路径 28 vector<int> temp; 29 for(int i= 0;i < path_index;i++) 30 temp.push_back(path[i]); 31 temp.push_back(sum);//叶节点这里才进入向量 32 answer.push_back(temp); 33 } 34 else{ 35 sum -= root->val; 36 path[path_index++] = root->val; 37 if(root->left != NULL) 38 dfs(root->left,sum,path_index); 39 if(root->right != NULL) 40 dfs(root->right,sum,path_index); 41 } 42 } 43 }; 44 int main(){ 45 TreeNode* treenode = new TreeNode(5); 46 TreeNode* left = new TreeNode(4); 47 treenode->left = left; 48 TreeNode* right = new TreeNode(8); 49 treenode->right = right; 50 51 Solution s; 52 vector<vector<int> > a = s.pathSum(treenode,9); 53 for(int i = 0;i < a.size();i++){ 54 std::vector<int> v = a[i]; 55 for(int j = 0;j < v.size();j++) 56 cout << v[j]<<" "; 57 cout <<endl; 58 } 59 60 }
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