PAT A1102 Invert a Binary Tree

Sample Input:

8
1 -
- -
0 -
2 7
- -
- -
5 -
4 6

Sample Output:

3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1

word	meaning
Max Howell	马克斯·豪威尔
indices	(index的复数)

思路 1：使用静态二叉树（含父节点）

findRoot：从0节点开始，不断向上遍历到父节点为空（f == -1）
从root开始，进行逆层序遍历，逆中序遍历（交换遍历左右子树的顺序即可）
层序遍历，直接输出每次出队的元素，中序遍历是递归，不好控制输出，故用一个数组存储结果，遍历后再格式化输出

code 1:

#include <stdio.h>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 110;
int n, root;
char tmpl, tmpr;
struct Node{int data, lc, rc, f;Node(){f = lc = rc = -1;}
}node[maxn];
int findRoot(){int idex = 0;while(node[idex].f != -1){idex = node[idex].f;}return idex;
}
void revLevelOrder(int root){queue<int> q;q.push(root);int cnt = 0;while(!q.empty()){int now = q.front();cnt++;printf("%d", node[now].data);if(cnt != n) printf(" ");q.pop();if(node[now].rc != -1) q.push(node[now].rc); if(node[now].lc != -1) q.push(node[now].lc);}printf("\n");
}
vector<int> in;
void revInOrder(int root){if(node[root].rc != -1) revInOrder(node[root].rc);in.push_back(root);if(node[root].lc != -1) revInOrder(node[root].lc);
}
void outputIn(){for(int i = 0; i < n; ++i){printf("%d", in[i]);if(i < n-1) printf(" ");}
}
int main(){scanf("%d", &n);for(int i = 0; i < n; ++i){getchar();    //!!!:Wrong 1: scanf读入char，要吸收掉上一轮结束时的\n scanf("%c %c", &tmpl, &tmpr);if(tmpl == '-') node[i].lc = -1;else{node[tmpl-'0'].f = i;node[i].lc = tmpl - '0';} if(tmpr == '-') node[i].rc = -1;else{node[tmpr-'0'].f = i;node[i].rc = tmpr - '0'; }node[i].data = i;}root = findRoot();revLevelOrder(root);revInOrder(root);outputIn();return 0;
}

不需要output控制inOrder的输出：

vector<int> in;
int cnt1 = 0;
void revInOrder(int root){if(root == -1) return;if(node[root].rc != -1) revInOrder(node[root].rc);printf("%d", root);cnt1++;if(cnt1 < n) printf(" ");if(node[root].lc != -1) revInOrder(node[root].lc);
}

TIPS: 遍历之前要判断结点是否为空
思路 2：
在输入时，就交换左右孩子，然后按正常的层序遍历和中序遍历输出
code 2:

#include <stdio.h>
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 110;
int n, root;
char tmpl, tmpr;
struct Node{int lc, rc; //!!! TIPS 1:不需要data，下标就是idex
}node[maxn];
bool has[maxn]; //找root
int cnt = 0;
void print(int root){//单独用一个print()控制输出格式 printf("%d", root);cnt++;if(cnt < n) printf(" ");else printf("\n");
}
int change(char c){return c == '-' ? -1 : c-'0';
}
int findRoot(){for(int i = 0; i < n; ++i){if(has[i] == false)return i;}
}
void levelOrder(int root){queue<int> q;q.push(root);while(!q.empty()){int now = q.front();print(now);q.pop();if(node[now].lc != -1) q.push(node[now].lc);if(node[now].rc != -1) q.push(node[now].rc); }cnt = 0;
}
void inOrder(int root){if(root == -1) return;if(node[root].lc != -1) inOrder(node[root].lc);print(root);if(node[root].rc != -1) inOrder(node[root].rc);
}
int main(){scanf("%d", &n);for(int i = 0; i < n; ++i){scanf("%*c%c %c", &tmpl, &tmpr);    //!!!:TIPS 2:"%*c"可以在scanf()中吸收一个字符 int lchild = change(tmpl), rchild = change(tmpr);//！！！输入时交换左右孩子 node[i].rc = lchild;if(lchild != -1) has[lchild] = true;node[i].lc = rchild; if(rchild != -1) has[rchild] = true;}root = findRoot();levelOrder(root);inOrder(root);return 0;
}

TIPS 2: "%*c" 可以在scanf()中吸收一个字符
思路 3：
先用后序遍历交换左右孩子，再层序和中序遍历
code 3:

void postOrder(int root){if(root == -1) return;if(node[root].lc != -1) postOrder(node[root].lc);if(node[root].rc != -1) postOrder(node[root].rc);swap(node[root].lc, node[root].rc);
}

T2 code:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 15;
struct Node{int lc, rc;
}node[maxn];
bool HasFather[256];
int FindRoot(int n){for(int i = 0; i < n; ++i){if(HasFather[i + '0'] == false) return i;}
}
//bool inq[maxn];
void RevLevel(int root, int n){int cnt = 0;queue<int> q;q.push(root);
//  inq[root] = true;  //层序遍历不需要inq[] while(!q.empty()){int now = q.front();printf("%d", now);if(++cnt < n) printf(" ");q.pop();if(node[now].rc != -1) q.push(node[now].rc);if(node[now].lc != -1) q.push(node[now].lc);}printf("\n");
}
int cntIn = 0;
void RevIn(int root, int n){if(root == -1) return;RevIn(node[root].rc, n);    printf("%d", root);if(++cntIn < n) printf(" ");RevIn(node[root].lc, n);
}int main(){int n;scanf("%d", &n);for(int i = 0; i < n; ++i){char tl, tr;getchar();scanf("%c %c", &tl, &tr);node[i].lc = tl != '-' ? tl - '0' : -1;node[i].rc = tr != '-' ? tr - '0' : -1;HasFather[tl] = true;HasFather[tr] = true;}int root = FindRoot(n);RevLevel(root, n);RevIn(root, n);return 0;
}

T3 code: 数组存储二叉树，先逆中序遍历，将中序序列存入一个vector，同时用id控制将节点存入一个数组，在遍历数组即得到逆层序序列

#include <bits/stdc++.h>
using namespace std;
const int maxn = 110;
int seq[maxn];
struct Node
{int data, lc, rc;
}node[maxn];
int Change(char s[])
{return s[0] == '-' ? -1 : atoi(s);
}
vector<int> in;
void RevInorder(int id, int root)
{if(root == -1) return;RevInorder(id * 2, node[root].rc);in.push_back(root);seq[id] = root;RevInorder(id * 2 + 1, node[root].lc);
}
int main()
{int n;scanf("%d", &n);int r = (n - 1) * n / 2 - (n + 1);for(int i = 0; i < n; ++i){char lc[5], rc[5];scanf("%s %s", lc, rc);node[i].lc = Change(lc);node[i].rc = Change(rc);r -= (node[i].lc + node[i].rc);}memset(seq, -1, sizeof(seq));RevInorder(1, r);printf("%d", seq[1]);for(int i = 2; i < maxn; ++i){if(seq[i] != -1) printf(" %d", seq[i]);}printf("\n");for(int i = 0; i < n; ++i){printf("%d", in[i]);if(i < n-1) printf(" ");}return 0;
}

PAT A1102 Invert a Binary Tree相关推荐

PAT甲级Invert a Binary Tree 柳神层序遍历的思路值得借鉴
1102 Invert a Binary Tree (25分) The following is from Max Howell @twitter: Google: 90% of our engine ...
PAT甲级1102 Invert a Binary Tree：[C++题解]反转二叉树、递归
文章目录题目分析题目链接题目分析反转二叉树这道题目特别出名!!!,是因为Homebrew这款Mac上大火的软件的作者到google面试,就考了这道题.面试官说了下面这段话:你竟然连在白板上翻转 ...
PAT甲级——1102 Invert a Binary Tree (层序遍历+中序遍历)
本文同步发布在CSDN:https://blog.csdn.net/weixin_44385565/article/details/90577042 1102 Invert a Binary Tree ...
1102. Invert a Binary Tree (25)-PAT甲级真题
The following is from Max Howell @twitter: Google: 90% of our engineers use the software you wrote ( ...
PAT甲级-二叉树的遍历-1102 Invert a Binary Tree解题思路
1102 Invert a Binary Tree (25 分) 思路翻转二叉树后序遍历翻转即可,由于给出每个结点的左右儿子,所以这里用到二叉树的静态写法更加方便这里有个坑,bool数组初始化为 ...
1102 Invert a Binary Tree
题目来源:PAT (Advanced Level) Practice The following is from Max Howell @twitter: Google: 90% of our eng ...
PAT甲级1110 Complete Binary Tree：[C++题解]判断完全二叉树
文章目录题目分析题目链接题目分析分析: 按照层序的顺序将完全二叉树存在下标从1开始的数组中.如果是完全二叉树,会将数组中1 ~ n这些位置填满,最大下标就是n,如果最大下标大于n,说明中间有空 ...
1102 Invert a Binary Tree(甲级)
1102 Invert a Binary Tree (25分) The following is from Max Howell @twitter: Google: 90% of our engine ...
Invert a Binary Tree
The following is from Max Howell @twitter: Google: 90% of our engineers use the software you wrote ( ...

PAT A1102 Invert a Binary Tree

PAT A1102 Invert a Binary Tree

Sample Input:

Sample Output:

PAT A1102 Invert a Binary Tree相关推荐

最新文章

热门文章