将二叉树PreOrder, inOrder序列化后，再恢复树形结构

最新推荐文章于 2024-06-14 16:09:55 发布

leo524891010

最新推荐文章于 2024-06-14 16:09:55 发布

阅读量1.5k

点赞数

CC 4.0 BY-SA版权

本文链接：https://blog.youkuaiyun.com/algorithmengine/article/details/7908888

本文详细阐述了如何通过前序和中序遍历字符串序列化二叉树，并通过这些字符串获取原始树结构的过程。包括测试场景设计、算法实现与优化策略。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

/*
you are given a system of passing binary trees among 2 ppl

Step1: convert the tree to preorder and inorder strings
Step2:send those strings to the intended person
Step3:get back tree from the strings

whats your strategy of testing?write various test scenarios.---10 marks
*/

#include <iostream>

char pre[7];
char in[7];
int cnt = 0;

struct node
{
	char data;
	node* left;
	node* right;
	node(int eData, node* eLeft, node* eRight)
	{
		data = eData;
		left = eLeft;
		right = eRight;
	}
};

void preOrder(node* n)
{
	if (!n)
		return;

	pre[cnt++] = n->data;
	preOrder(n->left);
	preOrder(n->right);
};

void inOrder(node* n)
{
	if (!n)
		return;
	
	inOrder(n->left);
	in[cnt++] = n->data;
	inOrder(n->right);
};

node* restoreTree(int preLeft, int preRight, int inLeft, int inRight)
{
	if (preLeft < preRight)
	{
		char curData = pre[preLeft];
		int i = inLeft;
		for (; i <= inRight; i++)
		{
			if (in[i] == curData)
				break;
		}

		int num = i - inLeft;
		node* curNode = new node(curData, NULL, NULL);
		curNode->left = restoreTree(preLeft + 1, preLeft + num, inLeft, i - 1);
		curNode->right = restoreTree(preLeft + num + 1, preRight, i + 1, inRight);
		return curNode;
	}
	else if (preLeft == preRight)
		return new node(pre[preLeft], NULL, NULL);

};

int main()
{
	node* n6 = new node('G', NULL, NULL);
	node* n5 = new node('D', n6, NULL);
	node* n4 = new node('E', NULL, NULL);
	node* n3 = new node('B', n5, n4);
	node* n2 = new node('F', NULL, NULL);
	node* n1 = new node('C', n2, NULL);
	node* n0 = new node('A', n3, n1);
	
	cnt = 0;
	preOrder(n0);

	cnt = 0;
	inOrder(n0);

	node* root = restoreTree(0, 6, 0, 6);

	return 0;
}