hdu 1710 Binary Tree Traversals

根据一颗二叉树的先序遍历结果和中序遍历结果确定后序遍历结果。

先递归建树，后DFS。代码写了详细的注释~~~~

至今还不会用指针写数据结构。。只会用结构体模拟。。。

 

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int maxn = 1111;
struct abc
{
    int left, right, date;
}node[maxn];
int xian[maxn], zhong[maxn];
int fxian[maxn], fzhong[maxn];
int flag[maxn];
int ans[maxn];
int sum, uu;
void gettree(int ll, int rr, int fn)//ll和rr表示的是中序的区间
{
    int i, maxn = -99999999, minn = 99999999;
    //minn maxn为中序ll到rr这段区间在先序的区间
    for (i = ll; i <= rr; i++)
    {
        if (fxian[zhong[i]] > maxn) maxn = fxian[zhong[i]];
        if (fxian[zhong[i]] < minn) minn = fxian[zhong[i]];
    }
    
    //中序ll到rr这段区间的节点为对应的先序序列中的第一个
    node[fn].date = xian[minn];
    
    //这段区间的节点在中序的位置
    int yy = fzhong[xian[minn]];

    //构造左子树
    if (yy - ll != 0)
    {
        sum++;
        node[fn].left = sum;
        gettree(ll, yy - 1, sum);
    }
    
    //构造右子树
    if (rr - yy != 0)
    {
        sum++;
        node[fn].right = sum;
        gettree(yy + 1, rr, sum);
    }

}
void dfs(int wei)//后序遍历 先遍历左儿子 再遍历右儿子 最后遍历节点
{
    //遍历左儿子
    if (node[wei].left == -1) flag[node[wei].left] = 1;
    else
    {
        
        dfs(node[wei].left);
        if (flag[node[wei].left] == 0) ans[uu] = node[node[wei].left].date, uu++;
        flag[node[wei].left] = 1;
        
    }
    //遍历右儿子
    if (node[wei].right == -1) flag[node[wei].right] = 1;
    else
    {
        
        dfs(node[wei].right);
        if (flag[node[wei].right] == 0) ans[uu] = node[node[wei].right].date, uu++;
        flag[node[wei].right] = 1;
        
    }
    //遍历节点
    flag[wei] = 1;
    ans[uu] = node[wei].date, uu++;
    
}
int main()
{
    int n, i;
    while (~scanf("%d", &n))
    {
        sum = 1; uu = 0;
        memset(flag, 0, sizeof(flag));
        for (i = 0; i <= maxn - 10; i++) node[i].date = -1, node[i].left = -1, node[i].right = -1;
        for (i = 1; i <= n; i++) scanf("%d", &xian[i]);
        for (i = 1; i <= n; i++) scanf("%d", &zhong[i]);
        for (i = 1; i <= n; i++){fxian[xian[i]] = i; fzhong[zhong[i]] = i;}
        gettree(1, n, 1);
        dfs(1);
        for (i = 0; i < uu; i++)
        {
            if (i<uu-1) printf("%d ", ans[i]);
            else printf("%d\n", ans[i]);
        }
    }
    return 0;
}