题目
The following is from Max Howell @twitter:
Google: 90% of our engineers use the sofware you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck of. Now it’s your turn to prove that YOU CAN invert a binary tree!
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) which is the total number of nodes in the tree — and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node from 0 to N-1, and gives the indices of the lef and right children of the node. If the child does not exist, a “-” will be put at the position. Any pair of children are separated by a space.
Output Specification:
For each test case, print in the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.
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
题目分析
已知所有节点的左右子节点，求反转二叉树的中序和后序序列
解题思路
思路 01
- 输入时，将左右子节点对换，即可完成反转
- bfs广度优先遍历，输出层序序列
- 递归输出中序序列
思路 02
- 将节点关系按照输入保存
- 使用后序遍历递归进行二叉树反转（也可使用前序遍历递归进行二叉树反转）
- bfs广度优先遍历，输出层序序列
- 递归输出中序序列
Code
Code 01（最优）
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 10;
int nds[maxn][2];
int n,cnt;
bool flag[maxn];
void bfs(int root) {
queue<int> q;
q.push(root);
while(!q.empty()) {
int now = q.front();
q.pop();
printf("%d",now);
if(++cnt<n)printf(" ");
if(nds[now][0]!=-1)q.push(nds[now][0]);
if(nds[now][1]!=-1)q.push(nds[now][1]);
}
}
void inOrder(int nd){
if(nd==-1){//nds[nd][0]==-1&&nds[nd][1]==-1
return;
}
inOrder(nds[nd][0]);
printf("%d",nd);
if(++cnt<n)printf(" ");
inOrder(nds[nd][1]);
}
int main(int argc,char * argv[]) {
char f,r;
scanf("%d",&n);
for(int i=0; i<n; i++) {
scanf("%*c%c %c",&r,&f);
if(f=='-')nds[i][0]=-1;
else {
nds[i][0]=f-'0';
flag[nds[i][0]]=true;
}
if(r=='-')nds[i][1]=-1;
else {
nds[i][1]=r-'0';
flag[nds[i][1]]=true;
}
}
//find root
int k=0;
while(k<n&&flag[k])k++;
bfs(k);
printf("\n");
cnt=0;
inOrder(k);
return 0;
}
Code 02
#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 10;
struct node { // 二叉树的静态写法
int lchild, rchild;
} Node[maxn];
bool notRoot[maxn] = {false}; // 记录是否不是根结点，初始均是根结点
int n, num = 0; // n为结点个数，num为当前已经输出的结点个数
// print函数输出结点id的编号
void print(int id) {
printf("%d", id); // 输出id
num++; // 已经输出的结点个数加1
if(num < n) printf(" "); // 最后一个结点不输出空格
else printf("\n");
}
// 中序遍历
void inOrder(int root) {
if(root == -1) {
return;
}
inOrder(Node[root].lchild);
print(root);
inOrder(Node[root].rchild);
}
// 层序遍历
void BFS(int root) {
queue<int> q; //注意队列里是存地址
q.push(root); //将根结点地址入队
while(!q.empty()) {
int now = q.front(); //取出队首元素
q.pop();
print(now);
if(Node[now].lchild != -1) q.push(Node[now].lchild); //左子树非空
if(Node[now].rchild != -1) q.push(Node[now].rchild); //右子树非空
}
}
// 后序遍历，用以反转二叉树
//void postOrder(int root) {
// if(root == -1) {
// return;
// }
// postOrder(Node[root].lchild);
// postOrder(Node[root].rchild);
// swap(Node[root].lchild, Node[root].rchild); // 交换左右孩子
//}
// 前序遍历，用以反转二叉树
void preOrder(int root) {
if(root == -1) {
return;
}
swap(Node[root].lchild, Node[root].rchild); // 交换左右孩子
preOrder(Node[root].lchild);
preOrder(Node[root].rchild);
}
// 将输入的字符转换为-1或者结点编号
int strToNum(char c) {
if(c == '-') return -1; // “-”表示没有孩子结点，记为-1
else {
notRoot[c - '0'] = true; // 标记c不是根结点
return c - '0'; // 返回结点编号
}
}
// 寻找根结点编号
int findRoot() {
for(int i = 0; i < n; i++) {
if(notRoot[i] == false) {
return i; // 是根结点，返回i
}
}
}
int main() {
char lchild, rchild;
scanf("%d", &n); // 结点个数
for(int i = 0; i < n; i++) {
scanf("%*c%c %c", &lchild, &rchild); // 左右孩子
Node[i].lchild = strToNum(lchild);
Node[i].rchild = strToNum(rchild);
}
int root = findRoot(); // 获得根结点编号
// postOrder(root); // 后序遍历，反转二叉树
preOrder(root); // 前序遍历，反转二叉树
BFS(root); // 输出层序遍历序列
num = 0; // 已输出的结点个数置0
inOrder(root); // 输出中序遍历序列
return 0;
}