【PAT甲级】1020 Tree Traversals (25分)：树的创建、遍历

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

分析：树的创建、遍历

题意：给你后序、中序序列，输出层次遍历序列。

• 总记不清创建左右子树时，后序序列的下标

记住这个图

左子树记忆：假如左子树只有一个结点 postL=postL+leftNum-1=0

//左子树
Node->lchild=create(postL,postL+leftNum-1,inL,k-1);
//右子树
Node->rchild=create(postL+leftNum,postR-1,k+1,inR);

• 层序遍历的队列元素类型是node*不是node

代码

#include<iostream>
#include<queue>

using namespace std;

const int maxN=35;
int post[maxN],in[maxN];
int N;
struct node {
	int data;
	node* lchild;
	node* rchild;
};

//用后序和中序序列构造二叉树
node* create(int postL,int postR,int inL,int inR) {
	if(postL>postR) {//递归边界
		return NULL;
	}
	node* Node=new node();
	Node->data=post[postR];

	//在中序序列中，查找in[k]==post[postR]的节点
	int k=inL;
	for(; k<=inR; k++) {
		if(in[k]==post[postR])
			break;
	}
	int leftNum=k-inL;//左子树的结点个数

	//左子树
	Node->lchild=create(postL,postL+leftNum-1,inL,k-1);
	//右子树
	Node->rchild=create(postL+leftNum,postR-1,k+1,inR);

	return Node;
}

//层序遍历
int num=0;
void levelTra(node* root) {
	queue<node*> que;
	que.push(root);

	while(!que.empty()) {
		node* top=que.front();
		que.pop();
		printf("%d",top->data);
		num++;
		if(num<N) printf(" ");

		if(top->lchild!=NULL) que.push(top->lchild);
		if(top->rchild!=NULL) que.push(top->rchild);
	}
}

int main() {
	scanf("%d",&N);
	for(int i=0; i<N; i++) {
		scanf("%d",&post[i]);
	}
	for(int i=0; i<N; i++) {
		scanf("%d",&in[i]);
	}
	//构造二叉树
	node* root=create(0,N-1,0,N-1);
	//输出层序序列
	levelTra(root);


	return 0;
}