1020 Tree Traversals
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
解答
#include <bits/stdc++.h>
using namespace std;
const int max_n = 31;
struct node
{
int data;
node *lchild;
node *rchild;
};
int in[max_n], post[max_n];
int N;
node *create(int inL, int inR, int postL, int postR) //创建二叉树
{
if (postL > postR)
return nullptr;
//构建根节点
node *root = new node;
root->data = post[postR];
int k; //根节点在中序的位置
for (k = inL; k <= inR; k++)
if (in[k] == post[postR])
break;
int num_left = k - inL; //左子树结点的个数
root->lchild = create(inL, k - 1, postL, postL + num_left - 1);
root->rchild = create(k + 1, inR, postL + num_left, postR - 1);
return root;
}
int num = 0; //已输出的结点的个数
void BFS(node *root)
{
queue<node *> q;
q.push(root);
while (!q.empty())
{
node *temp = q.front();
printf("%d", temp->data);
q.pop();
num++;
if (num < N)
printf(" ");
if (temp->lchild != nullptr)
q.push(temp->lchild);
if (temp->rchild != nullptr)
q.push(temp->rchild);
}
}
int main()
{
scanf("%d", &N);
for (int i = 0; i < N; i++)
scanf("%d", &post[i]);
for (int i = 0; i < N; i++)
scanf("%d", &in[i]);
node *root = create(0, N - 1, 0, N - 1);
BFS(root);
return 0;
}
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