1020 Tree Traversals (25 point(s))

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本文介绍了一种算法，该算法可以根据二叉树的中序和后序遍历序列，重建二叉树并输出层次遍历序列。通过递归创建树结构，并利用队列进行层次遍历，实现了从给定序列到层次遍历序列的转换。

1020 Tree Traversals (25 point(s))

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

根据中序遍历和后序遍历得出层次遍历。

关键点：

1. 重建树，体会createTree()函数；

2. 采用队列的层次遍历。

注意点：

1. C++11引入nullptr作为空指针。C++11以下支持NULL（视为0）。尽量使用nullptr。

2. createTree函数注意终止条件（即某子树不存在，left>right，返回空指针）。

#include<iostream>
#include<queue>
using namespace std;
int post[30];
int in[30];
struct Node{
	int data;
	Node* left;
	Node* right; 
	Node(int d):data(d),left(NULL),right(NULL){ }
};  
Node* createTree(int root,int left,int right){
	//根据post[]的root位置的元素和in[]的left到right区间的元素构建树 
	if(left>right) return NULL;
	Node* r = new Node(post[root]);
	int idx = left;
	while(in[idx]!=post[root]){
		idx++;
	}//idx是在中序遍历中根节点的位置 
	r -> left = createTree(root-right+idx-1,left,idx-1);
	r -> right = createTree(root-1,idx+1,right);
	return r;//必须返回当前的根节点 
}
void levelOrder(Node* root){
	queue<Node*> q;
	q.push(root);
	bool isFirst=true;
	while(!q.empty()){
		Node* front = q.front();
		q.pop();
		if(isFirst){
			cout<<front->data;
			isFirst = false;
		}
		else cout<<" "<<front->data;
		if(front->left!=NULL) q.push(front->left);
		if(front->right!=NULL) q.push(front->right);
	}
}
int main(void){
	int n;cin>>n;
	for(int i=0;i<n;i++) cin>>post[i];
	for(int i=0;i<n;i++) cin>>in[i];
	Node* root = createTree(n-1,0,n-1);
	levelOrder(root);
	return 0;
}