1020 Tree Traversals (25 分)
标签：front key layout post ram i++ flex htm cond
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
#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：front key layout post ram i++ flex htm cond
原文地址：https://www.cnblogs.com/cunyusup/p/10801579.html
