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<stdio.h> #include<string.h> 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<n) printf("%d ", ans[i]); else printf("%d ", ans[i]); } } return 0; }