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  • 1020. Tree Traversals (25)

    题目连接:https://www.patest.cn/contests/pat-a-practise/1020

    原题如下:

    Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

    Input Specification:

    Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

    Output Specification:

    For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

    Sample Input:
    7
    2 3 1 5 7 6 4
    1 2 3 4 5 6 7
    
    Sample Output:
    4 1 6 3 5 7 2
    这道题是要求利用中序和后序写出层序。不用建树即可,利用递归。我写的还是有些繁琐了,主要是中间那块不太确定,求稳了些……
    大致的思路就是先在后序中找到根,存入数组中,然后递归的解决左右两边,递归终止条件是参数中左边的序列大于右边。
    代码如下:
     1 #include<stdio.h>
     2 #define MAXN 65535
     3 int lastnum=0,N,PostOrder[MAXN],InOrder[MAXN];
     4 int LevelOrder[MAXN]={0};
     5 
     6 void Find(int Istart,int Iend,int Pstart,int Pend,int num)
     7 {
     8     if (Iend<Istart||Pend<Pstart)return ;
     9     else
    10     {
    11         LevelOrder[num]=PostOrder[Pend];
    12        // printf("%d ",PostOrder[Pend]);
    13         if (num>lastnum)lastnum=num;
    14         int i;
    15         for (i=Istart;i<=Iend;i++)
    16         {
    17             if (InOrder[i]==PostOrder[Pend])break;
    18         }
    19         int center,Lchild,Rchild;
    20         int LengthLeft,LengthRight;
    21         int LeftEnd,RightEnd;
    22 
    23         center=i;
    24 
    25         LengthLeft=center-Istart;  //中序中的左边长度
    26         LengthRight=Iend-center;   //中序中的右边长度
    27 
    28         Lchild=num*2+1;  //左子树序号
    29         Rchild=num*2+2;  //右子树序号
    30 
    31         LeftEnd=Pstart+LengthLeft-1;  //后序中左半边的最右边界
    32         RightEnd=LeftEnd+LengthRight; //后序中右半边的最右边界
    33 
    34         Find(Istart,center-1,Pstart,LeftEnd,Lchild);
    35         Find(center+1,Iend,LeftEnd+1,RightEnd,Rchild);
    36 
    37     }
    38 }
    39 int main()
    40 {
    41     int i;
    42     scanf("%d",&N);
    43     for (i=0;i<N;i++)scanf("%d",&PostOrder[i]);
    44     for (i=0;i<N;i++)scanf("%d",&InOrder[i]);
    45 
    46     Find(0,N-1,0,N-1,0);
    47 
    48     //printf("%d
    ",lastnum);
    49     int flag=0;
    50     for (i=0;i<=lastnum;i++)
    51     {
    52         if (LevelOrder[i]!=0)
    53         {
    54            // printf("%d ",i);
    55             if (flag==0){printf("%d",LevelOrder[i]);flag=1;}
    56            else printf(" %d",LevelOrder[i]);
    57         }
    58     }
    59 
    60     return 0;
    61 }

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  • 原文地址:https://www.cnblogs.com/wuxiaotianC/p/6361751.html
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