标签：case null stream tree ecif std cout sam osi
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.
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.
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.
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
4 1 6 3 5 7 2基本思路：
1、因为中序遍历和后序遍历可以唯一确定一棵二叉树，所以首先构造处这颗树，然后进行层序遍历即可
2、建立二叉树的时候要判断左右子树长度为0，并且要注意参数的值传递。具体如何做可参考代码。
#include<iostream> #include<sstream> #include<algorithm> #include<map> #include<cmath> #include<cstdio> #include<string.h> #include<queue> #include<vector> using namespace std; struct Node { int data; Node *lchild,*rchild; }; void CreateTree(int inOrder[],int iL,int iR, int postOrder[],int pI,int pR,Node *&root) { int temp=0; while(inOrder[temp]!=postOrder[pR]) temp++; int leftLength=temp-iL; int rightLength=iR-temp; if(root==nullptr) { root =new Node(); root->data=postOrder[pR]; } if(leftLength>0) CreateTree(inOrder,iL,temp-1,postOrder,pI,pR-rightLength-1,root->lchild); if(rightLength>0) CreateTree(inOrder,temp+1,iR,postOrder,pI+leftLength,pR-1,root->rchild); } int a[31]; void level(Node *root) { queue<Node*> q; q.push(root); int i=0; while(!q.empty()) { Node *temp=q.front(); q.pop(); a[i++]=temp->data; if(temp->lchild) q.push(temp->lchild); if(temp->rchild) q.push(temp->rchild); } } int main() { int n; cin>>n; int postOrder[n],inOrder[n]; for(int i=0; i<n; i++) cin>>postOrder[i]; for(int i=0; i<n; i++) cin>>inOrder[i]; Node *root=nullptr; CreateTree(inOrder,0,n-1,postOrder,0,n-1,root); level(root); if(n>0) cout<<a[0]; for(int i=1;i<n;i++) cout<<" "<<a[i]; return 0; }
标签：case null stream tree ecif std cout sam osi
原文地址：https://www.cnblogs.com/zhanghaijie/p/10228464.html
