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本帖最后由 糖逗 于 2020-5-8 17:28 编辑
题目描述:
- 输入一棵二叉树和一个整数,打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。
-  
- 示例:
- 给定如下二叉树,以及目标和 sum = 22,
- 5
- / \
- 4 8
- / / \
- 11 13 4
- / \ / \
- 7 2 5 1
- 返回:
- [
- [5,4,11,2],
- [5,8,4,5]
- ]
-  
- 提示:
- 节点总数 <= 10000
- 来源:力扣(LeetCode)
- 链接:https://leetcode-cn.com/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof
- 著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
复制代码
- #include<iostream>
- #include <malloc.h>
- #include <vector>
- #include <math.h>
- #include <queue>
- #include<vector>
- using namespace std;
- struct TreeNode{
- int val;
- TreeNode* left;
- TreeNode* right;
- TreeNode(int x): val(x), left(NULL), right(NULL){
- }
- };
- TreeNode* CreateTree(vector<int> input){
- TreeNode* res = (TreeNode*)malloc(sizeof(TreeNode)*input.size());
- for(int i = 0; i < input.size(); i++){
- res[i].val = input[i];
- res[i].left = NULL;
- res[i].right = NULL;
- }
- for(int i= 0; i < input.size(); i++){
- if(2*i+1 < input.size()){
- res[i].left = &res[2*i+1];
- }
- if(2*i+2 < input.size()){
- res[i].right = &res[2*i+2];
- }
-
- }
- return res;
-
- }
- void middle(TreeNode* root, vector<vector<int> >& res, int left, int right, int depth){
- if(root == NULL) return;
- int insert = left + (right - left) / 2;
- res[depth][insert] = root->val;
-
- middle(root->left, res, left, insert - 1, depth + 1);
- middle(root->right, res, insert + 1, right, depth + 1);
- }
- int treeDepth(TreeNode* root){
- if(root == NULL || root -> val == 0) return 0;
- return max(treeDepth(root->left) + 1, treeDepth(root->right) + 1);
- }
-
- void PrintTree(TreeNode* root) {
- int depth = treeDepth(root);
- int width = pow(2, depth) - 1;
- vector<vector<int> > res(depth, vector<int>(width, 0));
- middle(root, res, 0, width - 1, 0);
- for(int i = 0; i < res.size(); i++){
- for(int j = 0; j < res[i].size();j++){
- if(res[i][j] == 0){
- cout << " ";
- }
- else{
- cout << res[i][j];
- }
-
- }
- cout << endl;
- }
- cout << "------------------" << endl;
- }
- void dfs(TreeNode* root, vector<vector<int> >& res, int sum, vector<int>& temp){
- temp.push_back(root -> val);
- if(sum == root -> val && !root->right && !root -> left){
- res.push_back(temp);
- }
- if(root -> right) dfs(root -> right, res, sum - root->val, temp);
- if(root -> left) dfs(root -> left, res, sum - root->val, temp);
- temp.pop_back();
- }
- vector<vector<int> > solution(TreeNode* root, int sum){
- vector<vector<int> > res;
- vector<int> temp;
- if(root == NULL) return res;
- dfs(root, res, sum, temp);
- return res;
- }
- int main(void){
- vector<int> input1;
- cout << "send numbers for the first tree" << endl;
- int number1;
- while(cin >> number1){
- input1.push_back(number1);
- }
- cin.clear();
- TreeNode* root = CreateTree(input1);
- PrintTree(root);
-
-
- int sum;
- cin >> sum;
- vector<vector<int> > res = solution(root, sum);
- for(int i = 0; i < res.size(); i++){
- for(int j = 0; j < res[0].size(); j++){
- cout << res[i][j] << " ";
- }
- cout << endl;
- }
-
- return 0;
- }
复制代码
注意事项:
1.回溯法
2.本地调试后未输出多个方案,但leetcode上通过了。 |
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