M - Binary Tree Traversals（建立二叉树）

末蓝、 2022-03-20 08:10 164阅读 0赞

### 题目描述： ###

A binary tree is a finite set of vertices that is either empty or consists of a root r and two disjoint binary trees called the left and right subtrees. There are three most important ways in which the vertices of a binary tree can be systematically traversed or ordered. They are preorder, inorder and postorder. Let T be a binary tree with root r and subtrees T1,T2.

In a preorder traversal of the vertices of T, we visit the root r followed by visiting the vertices of T1 in preorder, then the vertices of T2 in preorder.

In an inorder traversal of the vertices of T, we visit the vertices of T1 in inorder, then the root r, followed by the vertices of T2 in inorder.

In a postorder traversal of the vertices of T, we visit the vertices of T1 in postorder, then the vertices of T2 in postorder and finally we visit r.

Now you are given the preorder sequence and inorder sequence of a certain binary tree. Try to find out its postorder sequence.  
![在这里插入图片描述][13d014c9e5e3240eee00ab3d73dc79e8_v_1549619246]  
Input  
The input contains several test cases. The first line of each test case contains a single integer n (1<=n<=1000), the number of vertices of the binary tree. Followed by two lines, respectively indicating the preorder sequence and inorder sequence. You can assume they are always correspond to a exclusive binary tree.  
Output  
For each test case print a single line specifying the corresponding postorder sequence.  
Sample Input  
9  
1 2 4 7 3 5 8 9 6  
4 7 2 1 8 5 9 3 6  
Sample Output  
7 4 2 8 9 5 6 3 1  
题意：给出一棵树的前序和中序序列。求后序序列。  
分析： 利用这两个序列建树，然后直接后序遍历就可以了。不过建树有一定的难度，有一定的规律。  
可参考：[https://blog.csdn.net/qq\_43506138/article/details/86657176][https_blog.csdn.net_qq_43506138_article_details_86657176]

#include<bits/stdc++.h>
    typedef int Self;
    typedef struct BiTreeNode
    {
        Self digit;
        struct BiTreeNode *LeftNode;
        struct BiTreeNode *RightNode;
    }BiTree;
    Self PreOrder[1001],LenPre;
    Self InOrder[1001],LenIn;
    void CreateBiTree(BiTree **T,Self leftPre,Self rightPre,Self leftIn,Self rightIn)
    {
        if(leftIn>rightIn)
            return ;
        int i,pos=-1;
        for(i=leftIn;i<=rightIn;i++)
            if(InOrder[i]==PreOrder[leftPre])
                {
                    pos=i;break;
                }
        if(pos==-1)
             return;
        *T=(BiTree *)malloc(sizeof(BiTree));
        (*T)->LeftNode=NULL;
        (*T)->RightNode=NULL;
        (*T)->digit=InOrder[pos];
        //这里的传值过程就是规律
        CreateBiTree(&(*T)->LeftNode,leftPre+1,leftPre+pos-leftIn,leftIn,pos-1);
        CreateBiTree(&(*T)->RightNode,leftPre+pos-leftIn+1,rightPre,pos+1,rightIn);
    }
    Self digit[1001],k;
    void InTraver(BiTree *T)
    {
        if(T!=NULL)
        {
            InTraver(T->LeftNode);
            InTraver(T->RightNode);
            digit[k++]=T->digit;
        }
    }
    int main()
    {
        Self N;
        while(~scanf("%d",&N))
        {
            BiTree *Root;
            Root=NULL;
            LenPre=LenIn=N-1;
            for(int i=0;i<N;i++)
                scanf("%d",&PreOrder[i]);
            for(int i=0;i<N;i++)
                scanf("%d",&InOrder[i]);
            CreateBiTree(&Root,0,LenPre,0,LenIn);
            k=0;
            InTraver(Root);
            for(int i=0;i<k-1;i++)
                printf("%d ",digit[i]);
            printf("%d",digit[k-1]);
            printf("\n");
    
        }
    }

[13d014c9e5e3240eee00ab3d73dc79e8_v_1549619246]: https://vj.e949.cn/13d014c9e5e3240eee00ab3d73dc79e8?v=1549619246
[https_blog.csdn.net_qq_43506138_article_details_86657176]: https://blog.csdn.net/qq_43506138/article/details/86657176