The order of a Tree （前序遍历）-优快云博客

博客围绕二叉搜索树展开，指出其形状与插入键的顺序密切相关。任务是给定树的插入顺序，找出能生成相同形状树且字典序最小的插入顺序。思路是按先根、再左、再右的前序遍历顺序，根据数据建树并输出。

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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.

Input

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.

Output

One line with n intergers, which are the order of a tree that generate the same tree with the least lexicographic.

Sample Input

4

1 3 4 2

Sample Output

1 3 2 4

题意：n个数可以构成一颗二叉搜索树，然后寻找字典序最小的插入顺序，来构成一颗相同的树。

思路：二叉搜索树的左儿子小于父节点，右儿子大于父节点。我们想要字典序最小的插入顺序且构成的树和标准树一样，那么根就需要一样，因此要按照先根（保证树是一样的）再左，再右（字典序最小）。而这种格式就是二叉搜索树的前序遍历顺序。因此我们只需要根据数据建树，按照前序遍历输出即可。

代码如下：

#include<cstdio>
#include<cstring>
struct node
{
    int val;
    node *lch,*rch;
};
int flag;
node *insert(node *root,int x) //插入每个节点
{
    if(root==NULL)
    {
        node *q = new node;
        q->val=x;
        q->lch=q->rch=NULL;
        return q;
    }
    if(x<root->val)
        root->lch=insert(root->lch,x);
    else
        root->rch=insert(root->rch,x);
    return root;
}
void print(node *root)  //前序遍历输出
{
    if(root!=NULL)
    {
        if(flag)  //格式化输出  ，先根
            printf("%d",root->val);
        else
            printf(" %d",root->val);
        flag=0;
        print(root->lch); //左子树
        print(root->rch); //右子树
    }

}
int main()
{
    int n,x;
    while(~scanf("%d",&n))
    {
        flag=1;
        node *root=NULL;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&x);
            root=insert(root,x);//插入值
        }
        print(root);
        printf("\n");
    }
}