PAT甲級1020 Tree Traversals (25 分)

1020 Tree Traversals

作者: CHEN, Yue
單位: 浙江大學 
時間限制: 400 ms 
內存限制: 64 MB 
代碼長度限制: 16 KB

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

題意

給出一棵樹的後序遍歷和中序遍歷結果，讓你根據這兩個序列求出這棵樹的層序遍歷序列…

解答

#include <iostream>
#include<queue>
using namespace std;
struct Node{//樹節點
	int data;
	Node *lchild, *rchild;
	Node(int d) {
		data = d; lchild = rchild = NULL;
	}
};
int n,post[35],in[35];//分別存放中序和後續序列
int pos[35];
queue<Node*> q;
Node* tran(int a[],int b[],int postl,int postnum,int inl,int innum) {//遞歸求解
	if (postnum<=0)return NULL;
	int r=pos[postl+postnum-1];
	Node *temp = new Node(a[r]);
	temp->lchild = tran(a, b, postl, r-inl,inl,r-inl);
	temp->rchild = tran(a, b, postl+r-inl,innum-(r-inl)-1,r+1,innum-(r-inl)-1);
	return temp;
}
int main(){
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> post[i];
	for (int i = 0; i < n; i++) {
		cin >> in[i];
		for (int j = 0; j < n; j++) {
			if (in[i] == post[j])
				pos[j] = i;
		}
	}
	Node *root=tran(in,post,0,n,0,n);
	q.push(root);bool first=true;
	while (!q.empty()) {
		if (q.front()->lchild != NULL)
			q.push(q.front()->lchild);
		if (q.front()->rchild != NULL)
			q.push(q.front()->rchild);
        if(first){
            first=false;
        }else
            cout<<" ";
		cout << q.front()->data;
		q.pop();
	}
	return 0;
}

這個破題花了我好多時間，找了好久才找到問題所在：寫代碼的時候有一個括號的位置寫錯了，導致中間遞歸的時候出現了問題，檢查的時候一直想着這個思路沒有問題，萬萬沒想到是這麼一個小細節上的問題，以後需要引以爲戒！！！切記！！