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

#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 = ;
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 = ;
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 - , inl, k-);
root->right = create(postl + leftnum, postr - , k + , inr);
return root;
}
int main() {
cin >> n;
for (int i = ; i < n; i++) {
cin>>post[i];
}
for (int i = ; i < n; i++) {
cin >> in[i];
}
node* root = create(, n - , , n - );
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。

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