1086 Tree Traversals Again (25 分)

1086 Tree Traversals Again (25 分)

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.

1086 Tree Traversals Again (25 分)_第1张图片
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

Analysis

题目大意

对一个有序的二叉树进行中序遍历时,可以使用非递归遍历。这时需要用到栈,现在给你一个压栈与出栈序列,让你将这个二叉树重建起来,并输出其后序遍历序列。

解析

数字先后压入的序列是二叉树的先序序列,数字出栈的序列是中序序列。根据两个序列就可以重建二叉树。

可以看我之前的这篇文章:重建二叉树。

Code

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 

using namespace std;

struct Node {
	int data;
	Node *lchild, *rchild;

	Node () {
		lchild = NULL;
		rchild = NULL;
	}
};

int n, cnt = 0, cur = 0;
vector pre, mid;
stack stk;

Node *root = new Node;

int findRootIndex(int num) {
	for(int i = 0; i < n; i++) {
		if (mid[i] == num) return i;
	}

	return -1;
}

// 使用先序序列与中序序列重建二叉树到根root上
Node* createTree(int left, int right) {
	if (left > right) return NULL;

	int rt = pre[cur];
	cur++;

	int rtIndex = findRootIndex(rt);

	Node *t = new Node;
	t->data = rt;

	if(left != right) {
		t->lchild = createTree(left, rtIndex - 1);
		t->rchild = createTree(rtIndex+1, right);
	}

	return t;
}

void postPrint(Node *r) {
	if (r == NULL) return;

	postPrint(r->lchild);

	postPrint(r->rchild);

	if (cnt != 0) printf(" ");
	if (cnt == 0) cnt = 1;

	printf("%d", r->data);
}

int main () {
	scanf("%d", &n);

	string s;
	int tmp;
	ios::sync_with_stdio(false);

	for(int i = 0; i < n*2; i++) {
		cin >> s;
		if (s[1] == 'u') {
			cin >> tmp;
			// push
			stk.push(tmp);
			pre.push_back(tmp);
		} else {
			// pop
			mid.push_back(stk.top());
			stk.pop();
		}
	}

	root = createTree(0, n-1);
	postPrint(root);
	return 0;
}

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