HDU 3999 The order of a Tree

The order of a Tree

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 845 Accepted Submission(s): 461


[align=left]Problem Description[/align]
As we know,the shape of a binary search tree is greatly related to the order of keys we insert. To be precisely:
1. insert a key k to a empty tree, then the tree become a tree with
only one node;
2. insert a key k to a nonempty tree, if k is less than the root ,insert
it to the left sub-tree;else insert k to the right sub-tree.
We call the order of keys we insert “the order of a tree”,your task is,given a oder of a tree, find the order of a tree with the least lexicographic order that generate the same tree.Two trees are the same if and only if they have the same shape.

[align=left]Input[/align]
There are multiple test cases in an input file. The first line of each testcase is an integer n(n <= 100,000),represent the number of nodes.The second line has n intergers,k1 to kn,represent the order of a tree.To make if more simple, k1 to kn is a sequence of 1 to n.

[align=left]Output[/align]
One line with n intergers, which are the order of a tree that generate the same tree with the least lexicographic.

[align=left]Sample Input[/align]

4

1 3 4 2

[align=left]Sample Output[/align]

1 3 2 4
题目大意：简历一颗二叉排序树，然后先序遍历。

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <stack>
using namespace std;

typedef struct node
{
int data;
node *lchild;
node *rchild;
node()
{
lchild = rchild = NULL;
}
}TreeNode;

void CreateTree(TreeNode *&pRoot, int data)
{
if (pRoot == NULL)
{
pRoot = new TreeNode;
pRoot->data = data;
}
else
{
if (data > pRoot->data)
{
CreateTree(pRoot->rchild, data);
}
else
{
CreateTree(pRoot->lchild, data);
}
}
}

void PreOrder(TreeNode *pRoot)
{
int nCount = 0;
if (pRoot == NULL)
{
return;
}
stack<TreeNode*> Stack;
Stack.push(pRoot);
do
{
TreeNode *p = Stack.top();
Stack.pop();
if (nCount == 0)
{
printf("%d", p->data);
nCount++;
}
else
{
printf(" %d", p->data);
nCount++;
}
if (p->rchild != NULL)
{
Stack.push(p->rchild);
}
if (p->lchild != NULL)
{
Stack.push(p->lchild);
}

} while (!Stack.empty());
}

int main()
{
int n, num;
scanf("%d", &n);
TreeNode *pRoot = NULL;
for (int i = 0; i < n; i++)
{
scanf("%d", &num);
CreateTree(pRoot, num);
}
PreOrder(pRoot);
printf("\n");
return 0;
}