PAT Advanced 1127 ZigZagging on a Tree (30) [中序后序建树，层序遍历]

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

1. dfs建树（结点左右指针表示树）
2. bfs遍历树，并设置每个节点分层保存
3. 锯齿形打印

思路 02

1. 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）
2. bfs遍历树，设置每个节点的层级，并分层保存
3. 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;
}