1127 ZigZagging on a Tree (30 分)

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 left to right and right to left. 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

/**
本题题意：
给出一颗树的后序，中序遍历（可以推导出先序遍历构建出唯一树），
输出此唯一树的层序遍历不过是z字型输出， eg: 第一层 从右往 左输出，第二层从左往右输出
方法一：本题思路
利用结点下标索引进行排序，先将高度排序，（根节点位于第0层）偶数层从右往左（从大到小排序，） 奇数层从左往右（从小到大排序）
**/

/**
本题题意：
给出一颗树的后序，中序遍历（可以推导出先序遍历构建出唯一树）， 
输出此唯一树的层序遍历不过是z字型输出， eg: 第一层 从右往 左输出，第二层从左往右输出 
本题思路 
利用结点下标索引进行排序，先将高度排序，（根节点位于第0层）偶数层从右往左（从大到小排序，） 奇数层从左往右（从小到大排序） 
**/ 
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
struct Node{
	int h, index, data;	   // 
};
int n;
vector<int> post, in;
vector<Node> zlevel;
bool cmp(Node a, Node b){
	if(a.h != b.h)
		return a.h < b.h;
	else if(a.h % 2 == 0){ //当高度一致时，偶数层从右往左 （索引从大到小输出) 
		return a.index > b.index;
	}else{                 //奇数层，从左往右，（索引从 小 到 大 输出） 
		return a.index < b.index;
	}
}
void preOrder(int postright, int inleft, int inright,int index, int h){
	if(inleft > inright) return;
	int i = inleft;
	while(post[postright] != in[i]) i++;
//	Node node = Node();//注意此步是 Node 没有new  
//	node.data = post[postright];
//	node.h = h;
//	node.index = index;
	zlevel.push_back({h, index, post[postright]});
	preOrder(postright - (inright - i) -1, inleft, i - 1, index * 2 + 1, h + 1);
	preOrder(postright - 1, i + 1, inright, index * 2 + 2, h + 1);
}
int main(){
	scanf("%d", &n);
	in.resize(n); 
	post.resize(n);
	for(int i = 0; i < n; i++){
		scanf("%d", &in[i]); 
	}
	for(int i = 0; i < n; i++){
		scanf("%d", &post[i]);
	}
	preOrder(n - 1, 0, n - 1, 0, 0);
	sort(zlevel.begin(), zlevel.end(), cmp);
	for(int i = 0; i < n; i++){
		if(i != 0) 
			printf(" ");
		printf("%d", zlevel[i].data);
	}
	return 0;
} 

 

方法二：

           根据中序后序推出前序遍历通过链表建立树， 通过BFS 得到层次遍历，将结点按高度存放在result二维数组中

偶数高度 逆序输出 （从右向左）

奇数高度 正序输出（从左至右）

具体代码：

/**
思路 ： 建立一颗树 将每层的结点通过BFS层序遍历分别放入二维向量数组 result[i]中(存放了在高度为i的全部结点) 
	   当高度为偶数时输出 倒序输出 
	   当高度为奇数时输出 正序输出   
**/
#include<iostream>
#include<queue>
#include<vector>
using namespace std;
struct Node{
	Node *l, *r;
	int data, h;
};
int n, cnt = 0; //cnt用来保证最后输出的格式 
vector<int> post, in, result[35]; // 

queue<Node *> q;
Node *preOrder(Node *root, int postright, int inleft, int inright, int h){
	if(inleft > inright){
		return nullptr;
	}
	int i = inleft;
	while(post[postright] != in[i]) i++;
	root = new Node();
	root -> data = post[postright];
	root -> h    = h;
	root -> l = preOrder(root -> l, postright -(inright - i) - 1, inleft, i - 1, h + 1);
	root -> r = preOrder(root -> r, postright - 1, i + 1, inright, h + 1);
	return root;
}

void levelOrder(Node *root){
	q.push(root);
	while(!q.empty()){
		Node *n = q.front();
		q.pop();
	 	result[n->h].push_back(n->data);
		if(n->l != nullptr) 
			q.push(n->l);
		if(n->r != nullptr)
			q.push(n->r);
		}
} 

int main(){
    //k % 2 == 0表示从左至右遍历 k % 2 == 1 表示从右至左遍历 
	Node *root = nullptr;
	scanf("%d", &n);
	in.resize(n);
	post.resize(n);
	for(int i = 0; i < n; i++){
		scanf("%d", &in[i]);
	}
	for(int i = 0; i < n; i++){
		scanf("%d", &post[i]);
	}
	root = preOrder(root, n - 1, 0, n - 1, 0);	 //构建出一棵树 
	levelOrder(root); //进行层序遍历并就结点放入二维向量数组中 
	printf("%d", result[0][0]);
	for(int i = 1; i < n; i++){
		if(i % 2 == 1)
			for(int j = 0; j < result[i].size(); j++)
	            printf(" %d", result[i][j]);			
		if(i % 2 == 0)
			for(int j = result[i].size() - 1; j >= 0; j--)
				printf(" %d", result[i][j]);
	}
	return 0;
}

当然也可以不用链表 静态用数组表示树：

#include<iostream>
#include<vector>
#include<queue>
using namespace std;
struct Node{
	int lindex, rindex, h, index;
}node[35];
int n;
queue<Node> q; 
vector<int> in, post, result[35];
int preOrder(int postright, int inleft, int inright, int h){
	if(inleft > inright) return -1;
	int i = inleft;
	while(post[postright] != in[i]) i++;
	node[postright].index = postright;
	node[postright].h = h;
	node[postright].lindex = preOrder(postright - (inright - i) - 1, inleft, i - 1, h + 1);
	node[postright].rindex = preOrder(postright - 1, i + 1, inright, h + 1);
	return postright;
}

void levelBFS(int postright){
	q.push(node[postright]);
	while(!q.empty()){
		Node n = q.front();
		result[n.h].push_back(post[n.index]);
		q.pop();
		if(n.lindex != -1){
			q.push(node[n.lindex]);
		}
		if(n.rindex != -1){
			q.push(node[n.rindex]);
		}
	}
	
}
int main(){
	int n;
	scanf("%d", &n);
	in.resize(n);
	post.resize(n);
	for(int i = 0; i < n; i++){
		scanf("%d", &in[i]);
	}
	for(int i = 0; i < n; i++){
		scanf("%d", &post[i]);
	}
	preOrder(n - 1, 0, n - 1, 0);
	levelBFS(n - 1);
	printf("%d", result[0][0]);
	for(int i = 1; i < n; i++){
		if(i % 2 == 1)
			for(int j = 0; j < result[i].size(); j++)
	            printf(" %d", result[i][j]);			
		if(i % 2 == 0)
			for(int j = result[i].size() - 1; j >= 0; j--)
				printf(" %d", result[i][j]);
	} 
	return 0;
}