1020 Tree Traversals (25 point(s))

1020 Tree Traversals (25 point(s))

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

Example:

#include<iostream>
#include<vector>
#include<queue>

using namespace std;

struct Order {
    int postBegin;
    int postEnd;
    int inBegin;
    int inEnd;
};

int main()
{
    int N;
    cin >> N;
    if(N == 0) return 0;
    vector<int> postOrder(N);
    vector<int> inOrder(N);
    for(int i = 0; i < N; i++) cin >> postOrder[i];
    for(int i = 0; i < N; i++) cin >> inOrder[i];
    queue<Order> Q;
    Q.push({0, N, 0, N});
    int count = 0;
    while(!Q.empty()) {
        Order &x = Q.front();
        int root = postOrder[x.postEnd-1];
        cout << (count++==0 ? "" : " ") << root;
        if(x.inEnd - x.inBegin > 1) {
            int i = x.inBegin;
            while(i < x.inEnd) {
                if(inOrder[i] == root) break;
                i++;
            }
            int left = i - x.inBegin;
            if(left > 0)
                Q.push({x.postBegin, x.postBegin+left, x.inBegin, x.inBegin+left});
            int right = x.inEnd - i - 1;
            if(right > 0)
                Q.push({x.postBegin+left, x.postEnd-1, x.inBegin+left+1, x.inEnd});
        }
        Q.pop();
    }
}

Ideas:

The last digit of the back-root traversal is the tree root, and the middle-root traversal is the left subtree before the root of the tree, and the right subtree is behind;

Judging that the number of sub-nodes is greater than 0, then enter the queue and wait for processing