PAT甲级7-4 1127 ZigZagging on a Tree (30分)

雨点打透心脏的1/2处 2023-02-27 12:21 42阅读 0赞

[1127 ZigZagging on a Tree (30分)][1127 ZigZagging on a Tree _30]  
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.

![在这里插入图片描述][20200717162249271.png]  
**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

题意：中后序建树 进行z字型输出，根据当前层号的奇偶性分别从左往右、从右往左遍历输出～

#include<bits/stdc++.h>
    using namespace std;
    int n,in[35],post[35],b[35][35]={ 0},k=0,ce=0;
    struct node{ 
    	int val,layer;
    	node *lc,*rc;
    };
    node *create(int inL,int inR,int postL,int postR){ 
    	if(postL>postR)return NULL;
    	node *root=new node;
    	root->val=post[postR];
    	int i;
    	for(i=inL;i<=inR;i++){ 
    		if(in[i]==post[postR])break;	
    	}
    	int leftnum=i-inL;
    	root->lc=create(inL,i-1,postL,postL+leftnum-1);
    	root->rc=create(i+1,inR,postL+leftnum,postR-1);	
    	return root;
    }
    
    void Layerorder(node *root){ 
    	queue<node*> q;//注意队列里存地址
    	root->layer = 0;//根结点的层号为1 
    	q.push(root);//将根结点地址入队
    	while(!q.empty()){ 
    		node *now = q.front();//取出队首元素
    		q.pop();
    		if(ce==now->layer){ 
    				k++; 
    		}else{ 
    		  ce=now->layer	;
    		  k=0;
    		}
    	    b[now->layer][k]=now->val;
    	   
    		if(now->lc != NULL){ 
    			now->lc->layer = now->layer + 1;//左孩子的层号为当前层号+1 
    			q.push(now->lc);//左子树非空
    		}
    		if(now->rc != NULL){ 
    			now->rc->layer = now->layer + 1;//右孩子的层号为当前层号+1 
    			q.push(now->rc);//右子树非空
    		}	 
    	} 
    	printf("%d",b[0][1]);
    	for(int i=1;i<=ce;i++){ 
    			if(i%2==0){ 
    				for(int j=34;j>=0;j--){ 
    			       if(b[i][j]>0)  printf(" %d",b[i][j]);
    				}
    			}else{ 
    				for(int j=0;j<35;j++){ 
    			       if(b[i][j]>0)  printf(" %d",b[i][j]);
    			     else break;
    		        }
    			}
    	}
    }
    int main(){ 
    	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);
        Layerorder(root);
    	return 0;
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70]

[1127 ZigZagging on a Tree _30]: https://pintia.cn/problem-sets/994805342720868352/problems/994805349394006016
[20200717162249271.png]: /images/20230209/4a4e222a3e22467d839b207838f00c3c.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70]: /images/20230209/b0e04a18b40f45cd9808a02f1c06ed4e.png