1127 ZigZagging on a Tree (30 分) 后序和中序遍历建树后输出Z字形层次遍历

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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.

![zigzag.jpg][]

### 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：**后序和中序遍历建树后输出Z字形层次遍历

**思路：**建树把结点的左右儿子存在l和r数组中，因为要z字形层次遍历输出，所以我先把深搜跑了一遍深度，知道每个结点在第几层，然后正常的用queue跑层次遍历，把每个结点层在他相应的深度的vector里，然后根据深度的奇偶跑Z字形。

#include<stdio.h>
    #include<string.h>
    #include<math.h>
    #include<iostream>
    #include<string>
    #include<sstream>
    #include<algorithm>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<vector>
    using namespace std;
    #define inf 0x3f3f3f3f
    #define LL long long
    int post[1005],in[1005];
    int l[1005],r[1005];
    int dfs(int postl,int postr,int inl,int inr)
    {
    	if(postl>postr) return -1;
    	int root=post[postr];
    	int pos=inl;
    	while(in[pos]!=root) pos++;
    	int cnt=pos-inl;
    	l[root]=dfs(postl,postl+cnt-1,inl,pos-1);
    	r[root]=dfs(postl+cnt,postr-1,pos+1,inr);
    	return root;
    }
    int le[1005],maxl;
    void level(int root,int h)
    {
    	if(root==-1)return;
    	le[root]=h;
    	maxl=max(maxl,h);
    	if(l[root]!=-1) level(l[root],h+1);
    	if(r[root]!=-1) level(r[root],h+1);
    }
    int main()
    {
    	int n,i,j;
    	memset(l,-1,sizeof l);
    	memset(r,-1,sizeof r);
    	scanf("%d",&n);
    	for(i=1;i<=n;i++)
    	scanf("%d",&in[i]);
    	for(i=1;i<=n;i++)
    	scanf("%d",&post[i]);
    	int root=post[n];
    	dfs(1,n,1,n);
    	level(root,1);
    	queue<int> q;
    	vector<int> v[1005]; 
    	q.push(root);
    	while(!q.empty())
    	{
    		int u=q.front();
    		q.pop();
    		v[le[u]].push_back(u);
    		if(l[u]!=-1)q.push(l[u]);
    		if(r[u]!=-1)q.push(r[u]);
    	}
    	printf("%d",root);
    	for(i=2;i<=maxl;i++)
    	{
    		if(i%2==0)
    		{
    			for(j=0;j<v[i].size();j++)
    			printf(" %d",v[i][j]);
    		}	
    		else
    		{
    			for(j=v[i].size()-1;j>=0;j--)
    			printf(" %d",v[i][j]);
    		}
    	}
    }

[zigzag.jpg]: https://imgconvert.csdnimg.cn/aHR0cHM6Ly9pbWFnZXMucHRhdXNlcmNvbnRlbnQuY29tLzMzN2NiZmIwLWE3YjItNDUwMC05NjY0LTMxOGU5ZmZjODcwZS5qcGc?x-oss-process=image/format,png