03-树3. Tree Traversals Again (25)

最新推荐文章于 2024-05-04 16:46:16 发布

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最新推荐文章于 2024-05-04 16:46:16 发布

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分类专栏： PAT.中国大学MOOC-陈越、何钦铭-数据结构基础习题集文章标签：浙大PAT MOOC C语言树的遍历非递归实现

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本文介绍了一种通过栈操作实现非递归二叉树遍历的方法，并给出了具体的代码实现，该方法能根据给定的栈操作序列生成唯一的二叉树并输出其后序遍历序列。

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03-树3. Tree Traversals Again (25)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

8000 B

判题程序

Standard

作者

CHEN, Yue

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.

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

#include <stdio.h>
struct Node{
	int tag;		//标记节点是第几次进栈
	int num;
};
//先序遍历对应进栈顺序，中序遍历对应出栈顺序；
//后序遍历与中序遍历不同的是节点出栈后要马上再入栈（tag做第二次入栈标记），等右儿子遍历完后再出栈;
//具体实现上，每次中序遍历的pop时，如果栈顶是标记过的节点（tag=2），循环弹出；如果没有标记过（tag=1），做标记，即弹出再压栈）
//栈顶tag=2的节点对应中序遍历中已弹出的节点；循环弹出后碰到的第一个tag=1的节点才对应中序遍历当前pop的节点
int main() {
	freopen("test.txt", "r", stdin);
	int n;
	scanf("%d", &n);
	int flag = 0;
	struct Node stack[30];	
	int size = 0;			//栈元素大小，指向栈顶的下一个位置
	for (int i = 0; i < 2 * n; ++i) {
		char s[10];
		scanf("%s", s);
		if (s[1] == 'u') {		//push
			scanf("%d", &stack[size].num);		//入栈
			stack[size].tag = 1;				//标记第一次入栈
			++size;
		}
		else {					//pop
			while (size > 0 && stack[size - 1].tag == 2) {	//循环弹出栈顶tag=2的节点
				if (flag)
					printf(" ");
				flag = 1;
				printf("%d", stack[--size].num);
			}
			if (size > 0)		//将中序遍历中应该要弹出的节点弹出再压栈，做标记即可
				stack[size - 1].tag = 2;
		}
	}
	while (size) {				//将栈中剩余节点依次弹出
		if (flag) {
			printf(" ");
		}
		flag = 1;
		printf("%d", stack[--size].num);
	}
	return 0;
}

题目链接：http://www.patest.cn/contests/mooc-ds/03-%E6%A0%913