题目
Suppose that all the keys in a binary tree are distinct positive integers. A unique binary tree can be determined by a given pair of postorder and inorder traversal sequences. And it is a simple standard routine to print the numbers in level-order. However, if you think the problem is too simple, then you are too naive. This time you are supposed to print the numbers in “zigzagging order” — that is, starting from the root, print the numbers level-by-level, alternating between lef to right and right to lef. For example, for the following tree you must output: 1 11 5 8 17 12 20 15.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<= 30), the total number of nodes in the binary tree. The second line gives the inorder sequence and the third line gives the postorder sequence. All the numbers in a line are separated by a space.
Output Specification:
For each test case, print the zigzagging sequence of the tree in a line. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.
Sample Input:
8
12 11 20 17 1 15 8 5
12 20 17 11 15 8 5 1
Sample Output:
1 11 5 8 17 12 20 15
题目分析
已知中序序列和后序序列，打印锯齿形序列（奇数层正序，偶数层倒序）
解题思路
思路 01
- dfs建树（结点左右指针表示树）
- bfs遍历树，并设置每个节点分层保存
- 锯齿形打印
思路 02
- dfs建树（二维数组表示树，每个数组含两个元素：左孩子节点和右孩子节点）
- bfs遍历树，设置每个节点的层级，并分层保存
- 锯齿形打印
Code
Code 01
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int maxn=30;
int n,pre[maxn],in[maxn],post[maxn];
struct node {
int data;
node * left;
node * right;
int depth;
};
vector<node*> result[30];
node * create(int inL,int inR,int postL,int postR) {
if(inL>inR)return NULL;
node * root = new node;
root->data=post[postR];
int k=inL;
while(k<inR&&in[k]!=post[postR])k++;
root->left=create(inL,k-1,postL,postR-(inR-k)-1);
root->right=create(k+1,inR,postR-(inR-k),postR-1);
return root;
}
void dfs(node * root) {
queue<node*> q;
root->depth=0;
q.push(root);
while(!q.empty()) {
node * now = q.front();
q.pop();
result[now->depth].push_back(now);
if(now->left!=NULL) {
now->left->depth=now->depth+1;
q.push(now->left);
}
if(now->right!=NULL) {
now->right->depth=now->depth+1;
q.push(now->right);
}
}
}
int main(int argc, char * argv[]) {
scanf("%d",&n);
for(int i=0; i<n; i++)scanf("%d",&in[i]);
for(int i=0; i<n; i++)scanf("%d",&post[i]);
node * root = create(0,n-1,0,n-1);
dfs(root);
printf("%d",result[0][0]->data);
for(int i=1;i<30;i++){
if(i%2==1){
for(int j=0;j<result[i].size();j++){
printf(" %d",result[i][j]->data);
}
}else{
for(int j=result[i].size()-1;j>=0;j--){
printf(" %d",result[i][j]->data);
}
}
}
return 0;
}
Code 02
#include <iostream>
#include <vector>
#include <queue>
using namespace std;
struct node {
int index;
int depth;
};
int n,tree[31][2],root;
vector<int> in,post,result[31];
void dfs(int &index, int inL, int inR, int postL, int postR) {
if(inL>inR)return;
index = postR; //当前root
//在中序序列中查找当前root
int k=inL;
while(k<inR&&in[k]!=post[postR])k++;
dfs(tree[index][0], inL, k-1, postL, postL+(k-inL)-1);
dfs(tree[index][1], k+1, inR, postL+(k-inL), postR-1);
}
void bfs() {
queue<node> q;
q.push(node {root,0});
while(!q.empty()) {
node temp = q.front();
q.pop();
result[temp.depth].push_back(post[temp.index]);
if(tree[temp.index][0]!=0)q.push(node{tree[temp.index][0],temp.depth+1});
if(tree[temp.index][1]!=0)q.push(node{tree[temp.index][1],temp.depth+1});
}
}
int main(int argc,char * argv[]) {
scanf("%d",&n);
in.resize(n+1),post.resize(n+1);
for(int i=1; i<=n; i++)scanf("%d",&in[i]);
for(int i=1; i<=n; i++)scanf("%d",&post[i]);
dfs(root,1,n,1,n);
bfs();
printf("%d",result[0][0]);
for(int i=1;i<31;i++){
if(i%2==1){
// 奇数行，正序
for(int j=0;j<result[i].size();j++){
printf(" %d",result[i][j]);
}
}else{
// 偶数行，逆序
for(int j=result[i].size()-1;j>=0;j--){
printf(" %d",result[i][j]);
}
}
}
return 0;
}