二叉树的应用与熟悉

一、二叉树的存储结构：

1.顺序存储结构：适用于完全二叉树

//-----二叉树的顺序表示------#define MAX_TREE_SIZE 100 //二叉树的最大结点数typedef TElemType SqBiTree[MAX_TREE_SIZE];//0号单元存储根结点SqBiTree bt; 

2.链式存储结构

//--------二叉树的二叉链表存储表示---------typedef struct BiNode{TElemType date;struct BiTNode *lchild,*rchild;//左右孩子指针}BiTNode,*BiTree; 

二、建立二叉树的二叉链表

status CreateBiTree(BiTree &T){ scanf(&ch); if(ch == ' ') T = null; else{ if(!(T = (BiTNode *)malloc(sizeof(BiTNode)))) exit(overflow); T->data = ch; CreateBiTree(T->lchild); CreateBiTree(T->rchild); } return ok;} 

poj2255

 

                                                                                 Tree Recovery

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 5290 Accepted: 3498

Description

Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital letters in the nodes. 
This is an example of one of her creations: 

                                               D
                                              / /
                                             /   /
                                            B     E
                                           / /     /
                                          /   /     /
                                         A     C     G
                                                    /
                                                   /
                                                  F

To record her trees for future generations, she wrote down two strings for each tree: a preorder traversal (root, left subtree, right subtree) and an inorder traversal (left subtree, root, right subtree). For the tree drawn above the preorder traversal is DBACEGF and the inorder traversal is ABCDEFG. 
She thought that such a pair of strings would give enough information to reconstruct the tree later (but she never tried it). 

Now, years later, looking again at the strings, she realized that reconstructing the trees was indeed possible, but only because she never had used the same letter twice in the same tree. 
However, doing the reconstruction by hand, soon turned out to be tedious. 
So now she asks you to write a program that does the job for her! 

Input

The input will contain one or more test cases. 
Each test case consists of one line containing two strings preord and inord, representing the preorder traversal and inorder traversal of a binary tree. Both strings consist of unique capital letters. (Thus they are not longer than 26 characters.) 
Input is terminated by end of file. 

Output

For each test case, recover Valentine's binary tree and print one line containing the tree's postorder traversal (left subtree, right subtree, root).

Sample Input

DBACEGF ABCDEFGBCAD CBAD

Sample Output

ACBFGEDCDAB

1.用到的一个函数：C++中substr函数用法

#include<string>#include<iostream>using namespace std;main(){string s("12345asdf");string a=s.substr(0,4);//获得字符串s中 从第0位开始的长度为4的字符串cout<<a<<endl;}输出结果为:1234 

2.问题分析：
已知树的先序和中序遍历序列求树后序遍历序列
先要根据先序和中序遍历序列建立树的二叉链表，再后续遍历就可以了
 
 
3.用到的知识点：
先序遍历序列的第一个值为该树的根结点，该值在中序遍历序列中把中序遍历序列一分为二，为该值的左子树和右子树
递归建立树的二叉链表
 
 
4.代码：
#include <string>#include <iostream>using namespace std;typedef struct Node{char data;struct Node *LTree,*RTree;}BinaryTree;/*PreCreateTree先序创建BinaryTree*/BinaryTree* Pre_Mid_CreateTree(string pre,string mid)  {BinaryTree* root = NULL;if(pre.length()>0){  root = new Node;root->data = pre[0];int index = mid.find(root->data);  root->LTree = Pre_Mid_CreateTree(pre.substr(1,index),mid.substr(0,index));root->RTree = Pre_Mid_CreateTree(pre.substr(index+1),mid.substr(index+1));  }  return root;  }void PostOrderTraverse(BinaryTree *head){if (head != NULL){PastOrderTraverse(head->LTree);PastOrderTraverse(head->RTree);printf("%c",head->data);}}int main(){BinaryTree *head=NULL;string pre,mid;while (cin>>pre>>mid){head = Pre_Mid_CreateTree(pre,mid);PostOrderTraverse(head);printf("/n");}}