Problem #
Source: [PAT 1020]
Description #
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 7Sample Output: #
4 1 6 3 5 7 2Solution #
后中序遍历构建二叉树并输出层次遍历
Code #
#include <iostream>
#include <queue>
#define MAXSIZE 31
using namespace std;
int in[MAXSIZE], post[MAXSIZE], n;
struct node
{
int v;
node *left, *right;
};
node *create(int inl, int inr, int postl, int postr)
{
if (postl > postr)
return NULL;
node *root = new node;
root->v = post[postr];
int i = 0;
for (i = inl; i <= inr; i++)
{
if (in[i] == post[postr])
break;
}
root->left = create(inl, i - 1, postl, postl + i - inl - 1);
root->right = create(i + 1, inr, postl + i - inl, postr - 1);
return root;
}
void travel()
{
queue<node *> q;
node *root = create(0, n-1, 0, n-1);
q.push(root);
bool flag = false;
while (!q.empty())
{
root = q.front();
q.pop();
if (flag)
cout << " ";
flag = true;
cout << root->v;
if(root->left!=NULL)q.push(root->left);
if(root->right!=NULL)q.push(root->right);
}
}
int main()
{
std::ios::sync_with_stdio(false);
std::cin.tie(0);
cin >> n;
for (int i = 0; i < n; i++)
cin >> post[i];
for (int i = 0; i < n; i++)
cin >> in[i];
travel();
return 0;
}
