1102 Invert a Binary Tree (25point(s))
The following is from Max Howell @twitter:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.
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 left 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
1、使用静态二叉树
2、后序遍历交换左右子树即为反转二叉树
#include<bits/stdc++.h>
using namespace std;
struct node{
int data;
int lchild,rchild;
}Node[110];
int rootFlag[110]={
false};
int n,num=0;
void print(int data){
printf("%d",data);
num++;
if(num!=n) printf(" ");
else printf("\n");
}
void inOrder(int root){
if(root==-1) return;
inOrder(Node[root].lchild);
print(root);
inOrder(Node[root].rchild);
}
void levelOrder(int root){
queue<int> q;
q.push(root);
while(!q.empty()){
int now=q.front();
q.pop();
print(now);
if(Node[now].lchild!=-1) q.push(Node[now].lchild);
if(Node[now].rchild!=-1) q.push(Node[now].rchild);
}
}
void postOrder(int root){
if(root==-1) return;
postOrder(Node[root].lchild);
postOrder(Node[root].rchild);
swap(Node[root].lchild,Node[root].rchild);
}
int sToNum(char c){
if(c=='-') return -1;
else{
rootFlag[c-'0']=true;
return c-'0';
}
}
int main(){
scanf("%d",&n);
char lchild,rchild;
for(int i=0;i<n;++i){
scanf("\n%c %c",&lchild,&rchild);//记得接收换行符,或者干脆直接用cin
Node[i].lchild=sToNum(lchild);
Node[i].rchild=sToNum(rchild);
}
int root=0;
for(int i=0;i<n;++i){
if(!rootFlag[i]){
root=i;
break;
}
}
postOrder(root);
levelOrder(root);
num=0;
inOrder(root);
}