OJ_判断两序列是否为同一二叉搜索树序列

题干

C++实现

1. 求出两棵树的中序和先序序列
2. 对比中序和先序序列是否相同

#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <string>
#include <queue>
using namespace std;

struct TreeNode {
	char data;
	TreeNode* left;
	TreeNode* right;
};

void InsertBST(TreeNode* &proot, int data) {
	TreeNode* pNew = new TreeNode();
	pNew->data = data;
	pNew->left = NULL;
	pNew->right = NULL;
	
	if (proot == NULL) {
		proot = pNew;
	}
	else {
		TreeNode* pCur = proot;//用于向下探索，插入结点
		TreeNode* pPre = NULL;//指向pCur的父亲
		while (1) {
			if (pCur->data > data) {
				pPre = pCur;
				pCur = pCur->left;
				if (pCur == NULL) {
					pPre->left = pNew;
					break;
				}
			}
			else {
				pPre = pCur;
				pCur = pCur->right;
				if (pCur == NULL) {
					pPre->right = pNew;
					break;
				}
			}
		}
	}
}

string preOrder(TreeNode* proot) {
	if (proot == NULL) {
		return "";
	}
	return proot->data + preOrder(proot->left) + preOrder(proot->right);
}

string inOrder(TreeNode* proot) {
	if (proot == NULL) {
		return "";
	}
	return inOrder(proot->left) + proot->data   + inOrder(proot->right);
}

int main() {
	int n;
	while (scanf("%d", &n) != EOF) {
		if (n == 0) {
			break;
		}
		char strArr[20] = { 0 };
		scanf("%s", strArr);
		TreeNode* proot1 = NULL;
		for (int i = 0; strArr[i] != '\0'; i++)
		{
			InsertBST(proot1, strArr[i]);
		}
		string inOrder1 = inOrder(proot1);
		string preOrder1 = preOrder(proot1);

		for (int i = 0; i < n; i++)
		{
			scanf("%s", strArr);
			TreeNode* proot2 = NULL;
			for (int j = 0; strArr[j] != '\0'; j++)
			{
				InsertBST(proot2, strArr[j]);
			}
			string inOrder2 = inOrder(proot2);
			string preOrder2 = preOrder(proot2);
			if (inOrder1 == inOrder2 && preOrder1 == preOrder2) {
				printf("YES\n");
			}
			else {
				printf("NO\n");
			}
		}
	}
	return 0;
}