A1020 Tree Traversals （25 分）

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 (≤), 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

 #include<cstdio>

 #include<queue>

 using namespace std;

 int post[],in[];

 struct node{

     int data;

     node *lchild;

     node *rchild;

 };

 node *create(int postL,int postR,int inL,int inR){

     if(postL>postR){

         return NULL;

     }

     node *root=new node;

     root->data=post[postR];

     int k;

     for(k=inL;k<=inR;k++){

         if(post[postR]==in[k]){

             break;

         }

     }

     int numleft=k-inL;

     root->lchild=create(postL,postL+numleft-,inL,k-);//这里经常出错

     root->rchild=create(postL+numleft,postR-,k+,inR);

     return root;

 }

 void LayerOrder(node*root,int k){

     if(root==NULL){

         return;

     }

     queue<node*> q;

     q.push(root);

     while(!q.empty()){

         node *now=q.front();

         q.pop();

         if(k!=){

             printf("%d ",now->data);

             k--;

         }else{

             printf("%d",now->data);

         }

         if(now->lchild!=NULL){

             q.push(now->lchild);

         }

         if(now->rchild!=NULL){

             q.push(now->rchild);

         }

     }

 }

 int main(){

     int n;

     scanf("%d",&n);

     for(int i=;i<n;i++){

         scanf("%d",&post[i]);

     }

     for(int i=;i<n;i++){

         scanf("%d",&in[i]);

     }

     node *p=create(,n-,,n-);

     LayerOrder(p,n);

     return ;

 }

Mist Note: 解决本题，本小编的思路是，先利用中序和后序序列重建二叉树，然后再层序遍历这棵树。

很多算法还是基础的，主要是在递归重建那里经常出问题。左子树和右子树的递归。

root->lchild=create(postL,postL+numleft-1,inL,k-);

root->rchild=create(postL+numleft,postR-,k+,inR);

红色部分经常出错，切记，我们在写代码的时候，经常是利用数组，而数组是以0开始的，而numleft是个数，你加之后，肯定需要-1，才符合定义，不要混淆。

本题还需要注意输出格式，最后层序遍历的最后一个节点，是不需要空格的。

巴特西

A1020 Tree Traversals （25 分）

Input Specification:

Output Specification:

Sample Input:

Sample Output:

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