1086 Tree Traversals Again (两种做法:模拟前序遍历和已知前序中序求树)

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

解题思路:

方法一:模拟前序遍历法

通过观察测试用例可以发现,这和先序遍历有很大的关联,每一次push,就是给当前结点赋值然后递归,每一次pop即返回上一个结点。这样我们可以写出递归遍历函数

nNode* XcreateTree(int arr[]) {
	if (arr[nindex] == -1) {
		nindex++;
		return nullptr;
	}
	nNode* newNode = new nNode(arr[nindex++]);
	if (newNode->lchild == nullptr) {
		newNode->lchild = XcreateTree(arr);
	}
	if (newNode->rchild == nullptr) {
		newNode->rchild = XcreateTree(arr);
	}
	return newNode;
}

我们定义一个数组,每次Push就把值存到数组中,每次pop就设置数组中的元素为-1,递归函数会根据数组中的值判断是继续遍历还是返回上一节点,AC代码如下:

#include <iostream>
#include <algorithm>
#include <vector>
#include <string>
#include <string.h>
using namespace std;
const int MAXN = 100;
int treeArr[MAXN];  //要用memset来设置
int N;
struct nNode
{
	int data;
	nNode *lchild, *rchild;
	nNode(int _data) {
		data = _data;
		lchild = rchild = nullptr;
	}
};
int nindex = 0;
nNode* XcreateTree(int arr[]) {
	if (arr[nindex] == -1) {
		nindex++;
		return nullptr;
	}
	nNode* newNode = new nNode(arr[nindex++]);
	if (newNode->lchild == nullptr) {
		newNode->lchild = XcreateTree(arr);
	}
	if (newNode->rchild == nullptr) {
		newNode->rchild = XcreateTree(arr);
	}
	return newNode;
}
//后序遍历输出结点
int num = 0;
void XpostOrder(nNode* root) {
	if (root->lchild != nullptr) {
		XpostOrder(root->lchild);
	}
	if (root->rchild != nullptr) {
		XpostOrder(root->rchild);
	}
	cout << root->data;
	num++;
	if (num < N) cout << " ";

}
int main() {
	scanf("%d", &N);
	memset(treeArr, -1, sizeof(treeArr));
	int nnIndex = 0;
	int rootData = 0;
	char str[5];
	int x;
	for (int i = 0; i < 2 * N; ++i) {
		scanf("%s", &str);
		if (strcmp(str, "Push") == 0) {
			scanf("%d", &x);
			treeArr[nnIndex++] = x;
		}
		else {
			treeArr[nnIndex++] = -1;
		}
	}
	nNode* root = XcreateTree(treeArr);
	XpostOrder(root);

	system("PAUSE");
	return 0;
}

方法二:已知前序中序建树法

通过观察测试用例可以发现,push的顺序为先序遍历顺序,pop的顺序为中序遍历顺序,这样的话,我们就可以把二叉树建立起来

AC代码如下:

#include <iostream>
#include <algorithm>
#include <string.h>
#include <string>
#include <stack>
using namespace std;
const int MAXN = 100;
struct mmNode
{
	int data;
	mmNode*lchild, *rchild;
	mmNode(int _data) {
		data = _data;
		lchild = rchild = nullptr;
	}
};
int preArrats[MAXN], inArrats[MAXN];

//套用模板即可
mmNode* createTree(int preL, int preR, int inL, int inR) {
	if (preL > preR) return nullptr;
	int curPreRoot = preArrats[preL];
	mmNode* newNode = new mmNode(curPreRoot);
	int k;
	for (k = inL; k <= inR; ++k) {
		if (curPreRoot == inArrats[k]) {
			break;
		}
	}
	int numk = k - inL;
	newNode->lchild = createTree(preL + 1, preL + numk, inL, k - 1);
	newNode->rchild = createTree(preL + numk + 1, preR, k + 1, inR);
	return newNode;
}
int N;
int num = 0;
void mmPostOrder(mmNode* root) {
	if (root == nullptr) return;
	if (root->lchild != nullptr) {
		mmPostOrder(root->lchild);
	}
	if (root->rchild != nullptr) {
		mmPostOrder(root->rchild);
	}

	cout << root->data;
	num++;
	if (num < N) cout << " ";

}
int main() {

	scanf("%d", &N);
	char str[5];
	int rootdata;
	stack<int> simulatorStack;
	int preIndex = 0, inIndex = 0;
	int x;
	for (int i = 0; i < 2 * N; ++i) {
		scanf("%s", &str);
		if (strcmp(str, "Push") == 0) {
			scanf("%d", &x);
			preArrats[preIndex++] = x;
			simulatorStack.push(x);
		}
		else {
			inArrats[inIndex++] = simulatorStack.top();
			simulatorStack.pop();
		}
	}

	mmNode* root = createTree(0, N - 1, 0, N - 1);
	mmPostOrder(root);

	return 0;
}