先建立一颗二叉搜索树，然后先序遍历一下，就可以出答案了。

我用了比较奇葩的写法。。。二叉搜索树结构体模拟了一下。先序遍历用的DFS。。。。。。

#include
     
      #include
      
       struct abc{    int left;    int right;    int date;}node[100010];int flag[100010];int ans[100010];int n, tot;void dfs(int wei){    if (node[wei].left != -1 && flag[node[wei].left] ==0)    {        flag[node[wei].left] = 1;        ans[tot] = node[node[wei].left].date;        tot++;        dfs(node[wei].left);    }    if (node[wei].right != -1 && flag[node[wei].right] == 0)    {        flag[node[wei].right] = 1;        ans[tot] = node[node[wei].right].date;        tot++;        dfs(node[wei].right);    }}int main(){    int i,  k;    while (~scanf("%d", &n))    {        memset(flag, 0, sizeof(flag));        for (i = 0; i <= n; i++) node[i].date = -1, node[i].left = -1, node[i].right = -1;        scanf("%d", &k);        tot = 1;        node[1].date = k;        int nodesum = 2;        for (i = 2; i <= n; i++)        {            scanf("%d", &k);            int nodexun = 1;            while (1)            {                if (k > node[nodexun].date)                {                    if (node[nodexun].right == -1)                    {                        node[nodexun].right = nodesum;                        node[nodesum].date = k;                        nodesum++;                        break;                    }                    else nodexun = node[nodexun].right;                }                else                {                    if (node[nodexun].left == -1)                    {                        node[nodexun].left = nodesum;                        node[nodesum].date = k;                        nodesum++;                        break;                    }                    else nodexun = node[nodexun].left;                }            }        }        ans[tot] = node[1].date;        tot++;        flag[1] = 1;        dfs(1);        for (i = 1; i <= n; i++)        {            if (i