1020. Tree Traversals

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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
#include<iostream>#include<queue>using namespace std;struct Node{Node* lchild;Node* rchild;int data;}Tree[32];int T;//分别保存后序和中序遍历序列 int PostOrder[32], InOrder[32];Node* create(){Tree[T].lchild = NULL;Tree[T].rchild = NULL;return &Tree[T++];}queue<Node*> que;//层序遍历Node *build(int s1, int e1, int s2, int e2){Node* root = create();root->data = PostOrder[e1];//后序遍历的最后一个节点是根节点 int rootIndex;for(int i = s2; i <= e2; i ++){//找到根节点在中序遍历序列中的位置 if(InOrder[i] == PostOrder[e1]){rootIndex = i;break;}}//递归的建立左子树与右子树 if(rootIndex != s2)root->lchild = build(s1, s1+rootIndex-s2-1, s2, rootIndex-1);if(rootIndex != e2)root->rchild = build(s1+rootIndex-s2, e1-1, rootIndex+1, e2);return root; }int main(){int n;while(cin>>n){for(int i = 0; i < n; i ++)cin>>PostOrder[i];for(int i = 0; i < n; i ++)cin>>InOrder[i];T = 0;Node* T = build(0, n-1, 0, n-1);que.push(T);while(!que.empty()){Node* tmp = que.front();if(tmp->lchild != NULL)que.push(tmp->lchild);if(tmp->rchild != NULL)que.push(tmp->rchild);que.pop();if(que.empty())cout<<tmp->data<<endl;elsecout<<tmp->data<<" ";}}return 0;} 


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