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 (≤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.
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 <stdio.h> #include <stdlib.h> #include <math.h> #include <algorithm> #include <iostream> #include <string.h> #include <queue> #include <string> #include <set> #include <map> using namespace std; const int maxn = 10000; int n; int post[maxn], in[maxn]; struct node { int data; node* left; node* right; }; void layerorder(node* root) { queue<node*> q; q.push(root); int count = 0; while (!q.empty()) { node* now = q.front(); q.pop(); printf("%d", now->data); count++; if (now->left != NULL) q.push(now->left); if (now->right != NULL) q.push(now->right); if (count != n) printf(" "); } } node* create(int postl, int postr, int inl, int inr) { if (postl > postr) { return NULL; } node* root = new node; root->data = post[postr]; int k; for (k = inl; k <= inr; k++) { if (in[k] == post[postr]) { break; } } int leftnum = k - inl; root->left = create(postl, postl + leftnum - 1, inl, k-1); root->right = create(postl + leftnum, postr - 1, k + 1, inr); return root; } int main() { cin >> n; for (int i = 0; i < n; i++) { cin>>post[i]; } for (int i = 0; i < n; i++) { cin >> in[i]; } node* root = create(0, n - 1, 0, n - 1); layerorder(root); system("pause"); }
注意点：考察基本的二叉树遍历，难点在递归上，想清楚了递归边界和递归式就简单了。二叉树的遍历及建树还需要巩固。
PAT A1020 Tree Traversals （25 分）
原文：https://www.cnblogs.com/tccbj/p/10392622.html