1020 Tree Traversals (25)（25 point(s)）

problem

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

tip

给出后序与中序遍历，要求层次遍历。

answer

#include<iostream>
#include<queue>
#define Max 33

using namespace std;

int n, back[Max], mid[Max];
//int tree[Max];
struct Node{
	int d;
	Node *l;
	Node *r;
};
Node *root;

void Print(){
	queue<Node*> q;
	q.push(root);
	while(!q.empty()){
		Node *t = q.front(); q.pop();
		if(!t->d) continue;
		if(!t->l && !t->r && q.empty()) cout<<t->d;
		else cout<<t->d<<" ";
		
		if(t->l) q.push(t->l);
		if(t->r) q.push(t->r);
	}
//	cout<<endl;
}

Node* DFS(int left, int right, int midLeft, int midRight){
	if(left > right) return NULL;
	
	Node *a = new Node();
	int num = back[right];
	int numIndex = -1;
	for(int i = midLeft; i <= midRight; i++){
		if(mid[i] == num) numIndex = i;
	}
	int lNum = numIndex - midLeft, rNum = midRight - numIndex;
	a->d = num;
	Print();
	a->l = DFS(left, left + lNum -1, numIndex - lNum, numIndex-1);
	a->r = DFS(right-rNum, right-1, numIndex+1, numIndex+rNum);
	return a;
}

int main(){
//	freopen("test.txt", "r", stdin);
	ios::sync_with_stdio(false);
	
	cin>>n;
	for(int i = 0; i < n; i++){
		cin>>back[i];
	}
	for(int i = 0; i < n; i++){
		cin>>mid[i];
	}
	root = new Node();
	root = DFS(0, n-1, 0, n-1);
	Print();
	return 0;
}

exprience

• 二叉树要多写几次，就熟悉了。