BUAA 2014级数据结构第五次上机二叉树之数组转换广义表

最新推荐文章于 2024-06-10 12:16:37 发布

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最新推荐文章于 2024-06-10 12:16:37 发布

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CC 4.0 BY-SA版权

文章标签：数据结构与算法

原文链接：http://www.cnblogs.com/zufezzt/p/4523251.html

本文介绍了一种构建二叉树的方法，并通过递归实现先序遍历输出结果。具体步骤包括初始化节点、读取输入数据构建树结构及进行先序遍历打印。

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按题意建立好二叉树，再按照先序遍历输出结果。

#include<cstdio>
#include<vector>
#include<queue>
#include<string.h>
#include<algorithm>
using namespace std;

struct node
{
    int left, right, date;

}node[1005];

int a[1005], flag[1005];

void dfs(int father)
{
    printf("%d", node[father].date);
    if (node[father].left != -1 || node[father].right != -1) printf("(");
    if (node[father].left != -1 && node[father].right == -1)
    {
        dfs(node[father].left);
        printf(")");
    }
    else if (node[father].left == -1 && node[father].right != -1)
    {
        printf(",");
        dfs(node[father].right);
        printf(")");
    }
    else if (node[father].left != -1 && node[father].right != -1)
    {
        dfs(node[father].left);
        printf(",");
        dfs(node[father].right);
        printf(")");
    }
    
}

int main()
{
    int n, i;
    while (~scanf("%d", &n))
    {
        int numm = 1;
        for (i = 0; i <= 1000; i++)
        {
            node[i].date = -1;
            node[i].left = -1;
            node[i].right = -1;
        }
        memset(flag, 0, sizeof(flag));
        for (i = 1; i <= n; i++) scanf("%d", &a[i]);
        int father = 1, jishu = 0;
        node[numm].date = a[1]; numm++;
        for (i = 2; i <= n; i++)
        {
            if (jishu == 0)
            {
                jishu++;
                if (a[i] != -1)
                {
                    node[father].left = numm;
                    node[numm].date = a[i];
                    numm++;
                }
            }
            else if (jishu == 1)
            {
                if (a[i] != -1)
                {
                    node[father].right = numm;
                    node[numm].date = a[i];
                    numm++;
                }
                jishu = 0;
                father++;
            }
        }
        dfs(1);
        printf("\n");
    }
    return 0;
}