|
马上注册,结交更多好友,享用更多功能^_^
您需要 登录 才可以下载或查看,没有账号?立即注册
x
本帖最后由 糖逗 于 2020-5-8 17:46 编辑
题目描述:
- 二叉搜索树中的两个节点被错误地交换。
- 请在不改变其结构的情况下,恢复这棵树。
- 示例 1:
- 输入: [1,3,null,null,2]
- 1
- /
- 3
- \
- 2
- 输出: [3,1,null,null,2]
- 3
- /
- 1
- \
- 2
- 示例 2:
- 输入: [3,1,4,null,null,2]
- 3
- / \
- 1 4
- /
- 2
- 输出: [2,1,4,null,null,3]
- 2
- / \
- 1 4
- /
- 3
- 进阶:
- 使用 O(n) 空间复杂度的解法很容易实现。
- 你能想出一个只使用常数空间的解决方案吗?
复制代码
- #include<iostream>
- #include <malloc.h>
- #include <vector>
- #include <math.h>
- #include <queue>
- #include<algorithm>
- 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 inorder(TreeNode* root, vector<int>& temp){
- if(root == NULL) return;
- inorder(root -> left, temp);
- if(root -> val != 0)temp.push_back(root -> val);
- inorder(root -> right, temp);
- }
- void rever(TreeNode* root, vector<int>& temp){
- if(root == NULL) return;
- rever(root -> left, temp);
- if(root -> val != 0){
- if(root -> val != temp.front()){
- root -> val = temp.front();
- temp.erase(temp.begin());
- }
- else{
- temp.erase(temp.begin());
- }
- }
-
- rever(root -> right, temp);
- }
- void solution(TreeNode* root) {
- if(root == NULL) return;
- vector<int> temp;
- inorder(root, temp);
- sort(temp.begin(), temp.end());
- for(int i =0; i < temp.size(); i++){
- cout << temp[i] << " ";
- }
- cout << endl;
- rever(root, temp);
-
- return;
-
- }
- int main(void){
- vector<int> input;
- int number;
- while(cin >> number){
- input.push_back(number);
- }
- TreeNode* root = CreateTree(input);
- PrintTree(root);
- solution(root);
- PrintTree(root);
-
- return 0;
- }
复制代码
注意事项:
1.中序遍历的二叉查找树的升序排列。
2.自己定义的树是用0代表缺失值的所以和leetcode上有些许不同(做了节点值是否为0的判断)。 |
|