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 7Sample Output:
4 1 6 3 5 7 2
分析:根据题目所给的后序遍历序列和中序遍历序列，构建二叉树，再输出该二叉树的前序遍历序列。
#include <iostream> #include <cstdio> #include <cmath> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; const int maxn=40; int post[maxn]; int in[maxn]; struct node { int data; node *lchild,*rchild; }; node * creat(int postL,int postR,int inL,int inR) { if(postL>postR) return NULL; node * root=new node; root->data=post[postR]; int index; for(index=inL;index<=inR;index++) { if(in[index]==post[postR]) { break; } } int numLeft=index-inL; root->lchild=creat(postL,postL+numLeft-1,inL,index-1); root->rchild=creat(postL+numLeft,postR-1,index+1,inR); return root; } int first_flag=0; void layerOrder(node * root) { queue<node *> ans; ans.push(root); while(!ans.empty()) { node * tmp=ans.front(); ans.pop(); if(first_flag!=0) { cout<<" "; } cout<<tmp->data; first_flag=1; if(tmp->lchild!=NULL) ans.push(tmp->lchild); if(tmp->rchild!=NULL) ans.push(tmp->rchild); } } int main() { int n; cin>>n; for(int i=0;i<n;i++) { cin>>post[i]; } for(int i=0;i<n;i++) { cin>>in[i]; } node * root; root=creat(0,n-1,0,n-1); layerOrder(root); return 0; }
原文：http://www.cnblogs.com/xiongmao-cpp/p/6430988.html