1020. Tree Traversals (25)

時間限制 400 ms 內(nèi)存限制 65536 kB 代碼長度限制 16000 B 判題程序 Standard 作者 CHEN, Yue

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 7 Sample Output: 4 1 6 3 5 7 2

分析:根據(jù)題目所給的后序遍歷序列和中序遍歷序列，構(gòu)建二叉樹，再輸出該二叉樹的前序遍歷序列。

#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; }

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轉(zhuǎn)載于:https://www.cnblogs.com/xiongmao-cpp/p/6430988.html

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