将二叉树PreOrder, inOrder序列化后，再恢复树形结构

/*
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;
}