1020 Tree Traversals (25 分)

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

已知中序、后序，建树，然后用队列层次遍历，保证最后一个不输出空格即可。

Code

#include<iostream>
#include<queue>
using namespace std;
int post[30], in[30];
int n;
struct Tnode{
	Tnode* left;
	Tnode* right;
	int val;
};
/*
根据后序、中序，递归建树
postL:后序序列左边, postR:后序序列右边, inL:中序序列左边, inR中序序列右边:
*/
Tnode* create(int postL, int postR, int inL, int inR){
	if(postL > postR)//后序序列长度小于0，递归结束 
		return NULL;
	Tnode* root = new Tnode;//新分配根节点，并赋值 
	root->val = post[postR];
	int k;
	for(k=inL; k<=inR; k++)
		if(in[k] == root->val)//在中序序列中找到根节点 
			break;
	int leftNum = k - inL;//中序根节点左子树包含节点数 
	root->left = create(postL, postL+leftNum-1, inL, k-1);
	root->right = create(postL+leftNum, postR-1, k+1, inR);
	return root;
}
/*层次遍历*/
void layerOrder(Tnode* root){
	int num = 0; 
	queue<Tnode*> q;//注意队列装的是结构体指针，* 
	q.push(root);
	while(!q.empty()){
		Tnode* now = q.front();q.pop();	//队首出队 
		printf("%d", now->val);		//访问该节点 
		if(num++ < n) printf(" ");//保证最后没有空格输出 
		if(now->left) q.push(now->left);//左子入队 
		if(now->right) q.push(now->right);//右子入队 
	}
}
int main(){
	cin >> n;
	//输入后序，中序 
	for(int i=0; i<n; i++)
		cin >> post[i];
	for(int i=0; i<n; i++)
		cin >> in[i];
	Tnode* root = create(0, n-1, 0, n-1);//建树 
	layerOrder(root);
	return 0;
}