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;

}

巴特西

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

题目