1020. Tree Traversals (25)

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

已知中序后序求层序，方法是根据后序找到根，再根据中序中根的位置找到右子树和左子树，题目中的树大概是下面这个样子

首先找到后序遍历中root肯定是最后一个元素4然后找到4在中序的位置，那么4右边的都为他的右子树4左边的都为他的左子树，然后下一个后序根为6，即右子树的根为6，6左边的为他左子树，右边的为他的右子树，依次递归。至于为什么是这样，因为后序遍历根始终是在其子树遍历完才会遍历到，而中序遍历则根在他左子树遍历好后经过根再遍历右子树。

前序遍历：根->左子树->右子树

中序遍历:  左子树->根->右子树

后序遍历:  左子树->右子树->根

/***已知后序中序求二叉树***/#include <iostream>#include <vector>#include <queue>using namespace std;typedef struct Tree{    int data;    Tree *left;    Tree *right;}Tree;Tree *root = NULL;vector<int> post;vector<int> in;int postindex;Tree *create(int left, int right){    if(left > right)    {        return NULL;    }    vector<int>::size_type i;    int rootdata = post[postindex];//取后序遍历的最后一个数据为根    postindex--;    Tree *p = new Tree();    p->data = rootdata;    for(i=0; i<in.size(); i++)//找到根在中序的位置    {        if(in[i] == rootdata)            break;    }    p->right = create(i+1, right);//中序根右边的数据为根的右子树，左边的为根的左子树，    p->left = create(left, i-1);    return p;}void levelorder()//层序遍历{    bool flag = true;    queue<Tree*> myque;    Tree *p;    p = root;    myque.push(p);    while(!myque.empty())    {        Tree *cur = myque.front();        myque.pop();        if(cur->left)            myque.push(cur->left);        if(cur->right)            myque.push(cur->right);        if(flag)        {            cout << cur->data;            flag = false;        }        else{            cout << " " << cur->data;        }    }}int main(){    int n, temp;    cin >> n;    for(int i=0; i<n; i++)    {        cin >> temp;        post.push_back(temp);    }    for(int i=0; i<n; i++)    {        cin >> temp;        in.push_back(temp);    }    postindex = post.size()-1;    root = create(0, n-1);    levelorder();    return 0;}