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

旧城等待， 2022-04-23 03:22 99阅读 0赞

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.

![30][]  
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;
    }

[30]: /images/20220323/2d121e1ede5f4fdd9503a2b016a6be27.png