首页 > 代码库 > Pat(Advanced Level)Practice--1086(Tree Traversals Again)

Pat(Advanced Level)Practice--1086(Tree Traversals Again)

2024-10-30 15:37:02   205人阅读

Pat1086代码

题目描述：

An inorder binary tree traversal can be implemented in a non-recursive way with a stack. For example, suppose that when a 6-node binary tree (with the keys numbered from 1 to 6) is traversed, the stack operations are: push(1); push(2); push(3); pop(); pop(); push(4); pop(); pop(); push(5); push(6); pop(); pop(). Then a unique binary tree (shown in Figure 1) can be generated from this sequence of operations. Your task is to give the postorder traversal sequence of this tree.

技术分享
Figure 1

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=30) which is the total number of nodes in a tree (and hence the nodes are numbered from 1 to N). Then 2N lines follow, each describes a stack operation in the format: "Push X" where X is the index of the node being pushed onto the stack; or "Pop" meaning to pop one node from the stack.

Output Specification:

For each test case, print the postorder traversal sequence of the corresponding tree in one line. A solution is guaranteed to exist. All the numbers must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:
6
Push 1
Push 2
Push 3
Pop
Pop
Push 4
Pop
Pop
Push 5
Push 6
Pop
Pop
Sample Output:
3 4 2 6 5 1

AC代码：二叉树重建+DFS
#include<cstdio>
#include<cstdlib>
#include<algorithm>

using namespace std;

char op[10];
int val,n,iter=0;
bool flag;

struct TreeNode{
	int num;
	TreeNode *left;
	TreeNode *right;
	TreeNode(int num):num(num),left(NULL),right(NULL){}
	~TreeNode();
};

TreeNode* buildTree(){
	if(iter>=n)
		return NULL;
	scanf("%s",op);
	TreeNode *root=NULL;
	if(op[1]=='u'){
		scanf("%d",&val);
		root=new TreeNode(val);
		++iter;
		root->left=buildTree();
		++iter;
		root->right=buildTree();
	}
	return root;
}

void DFS(TreeNode *root){
	if(root==NULL)
		return;
	DFS(root->left);
	DFS(root->right);
	if(flag){
		printf("%d",root->num);
		flag=false;
	}else{
		printf(" %d",root->num);
	}
}

int main(int argc,char *argv[]){
	scanf("%d",&n);
	n=n*2;
	TreeNode *root=buildTree();
	flag=true;
	DFS(root);
	printf("\n");
	return 0;
}

Pat(Advanced Level)Practice--1086(Tree Traversals Again)


联系
我们