[PAT-甲级]1020.Tree Traversals

1020. Tree Traversals (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

72 3 1 5 7 6 41 2 3 4 5 6 7

Sample Output:

4 1 6 3 5 7 2

解题思路：根据二叉树的后序和中序序列来建树，并打印这颗二叉树的层序遍历。递归的去建树。后序最后一个节点是根节点，该根节点将中序序列分为左右子树两部分。递归去做。

代码如下：

#include<stdio.h>#include<string.h>#include<queue>#include<algorithm>using namespace std;struct Node{int data;Node* leftChild;Node* rightChild;};int in[32], post[32];vector<int> level;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 countLeft = k - inL;root->leftChild = create(postL, postL+countLeft-1, inL, k-1);root->rightChild = create(postL+countLeft, postR-1, k+1, inR);return root;}void BFS(Node* root){queue<Node*> que;que.push(root);while(!que.empty()){Node *pre = que.front();que.pop();level.push_back(pre->data);if(pre->leftChild != NULL)que.push(pre->leftChild);if(pre->rightChild != NULL)que.push(pre->rightChild);}}int main(){freopen("D://input.txt", "r", stdin);int n;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);for(int i = 0; i < level.size(); i ++){if(i == 0)printf("%d", level[i]);elseprintf(" %d", level[i]);}return 0;}