1020. Tree Traversals (25)

Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

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 postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. 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:

7
2 3 1 5 7 6 4
1 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

思路：根据后序和中序遍历得出水平遍历顺序
 

  通过后序序列可以得到根和右子树的根的位子
 
 

  根据中序遍历序列可以得到根的左子树根节点离根节点的位移差
 
 

  这样就得到了根节点+左子根节点+右子根节点
 
 

  再利用一个队列即可得出水平遍历的序列
 
还可以用构建二叉树的方法来做，这样也比较通用，详细分析 狂戳这里

#include<iostream>
#include<queue>
#include<string.h>
using namespace std;
int *post,*visited;
const int N=31;
int hash_post[N],hash_in[N];
int n;
int right_size(int root){
	int i=0,size=0;
	for(i=hash_in[root]+1;i<=n-1;i++){
		if(visited[i]==false)
			size++;
		else
			break;
	}
	return size;
}
int main()
{
    cin>>n;
    post=new int[n];
	visited=new int [n];
    int i,tmp;
    for(i=0;i<n;i++){
        cin>>post[i];
		hash_post[post[i]]=i;
	}
    for(i=0;i<n;i++){
		cin>>tmp;
		hash_in[tmp]=i;
	}
	memset(visited,0,n*sizeof(int));
	queue<int> q;
	q.push(post[n-1]);
	int right=n-1;
	cout<<post[n-1];
	visited[hash_in[post[n-1]]]=true;
	while(!q.empty()){
		int root=q.front();
		int rsize=right_size(root);
		int left=hash_post[root]-rsize-1;//handle left sub tree root
		if(left>=0&&visited[hash_in[post[left]]]==false){
			int left_root=post[left];
			cout<<" "<<left_root;
			visited[hash_in[left_root]]=true;
			q.push(left_root);
		}
		int right=hash_post[root]-1;//handle right sub tree root
		if(right>=0&&visited[hash_in[post[right]]]==false){
			int right_root=post[right];
			cout<<" "<<right_root;
			visited[hash_in[right_root]]=true;
			q.push(right_root);
		}
		q.pop();
	}
    return 0;
}