PAT 1020 Tree Traversals
作者:互联网
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<bits/stdc++.h> using namespace std; typedef long long ll; struct node{ int data = -1; node *left = NULL; node *right = NULL; }; int n,cnt; int a[31]; int b[31]; node* createtree(int start,int end){ if(start > end) return NULL; int num = a[cnt--]; node *t = new(node); t->data = num; if(start == end){ return t;} int pos = -1; for(int i=start;i <= end;i++){ if(b[i] == num) pos = i; } t->right = createtree(pos+1,end); t->left = createtree(start,pos-1); return t; } int main(){ cin >> n; cnt = n-1; for(int i=0;i < n;i++) cin >> a[i]; for(int i=0;i < n;i++) cin >> b[i]; node *root = createtree(0,n-1); queue<node*> que; que.push(root); int count = 0; while(!que.empty()){ node *t = que.front();que.pop(); cout << t->data; if(count!=n-1){cout << " ";count++;} if(t->left)que.push(t->left); if(t->right)que.push(t->right); } return 0; }
创建节点要new,不然就会出错
标签:node,PAT,int,Traversals,Tree,start,que,end,line 来源: https://www.cnblogs.com/cunyusup/p/10801579.html