1020 Tree Traversals (25)（25 point(s)）

problem

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

tip

給出後序與中序遍歷，要求層次遍歷。

answer

#include<iostream>
#include<queue>
#define Max 33

using namespace std;

int n, back[Max], mid[Max];
//int tree[Max];
struct Node{
    int d;
    Node *l;
    Node *r;
};
Node *root;

void Print(){
    queue<Node*> q;
    q.push(root);
    while(!q.empty()){
        Node *t = q.front(); q.pop();
        if(!t->d) continue;
        if(!t->l && !t->r && q.empty()) cout<<t->d;
        else cout<<t->d<<" ";
        
        if(t->l) q.push(t->l);
        if(t->r) q.push(t->r);
    }
//  cout<<endl;
}

Node* DFS(int left, int right, int midLeft, int midRight){
    if(left > right) return NULL;
    
    Node *a = new Node();
    int num = back[right];
    int numIndex = -1;
    for(int i = midLeft; i <= midRight; i++){
        if(mid[i] == num) numIndex = i;
    }
    int lNum = numIndex - midLeft, rNum = midRight - numIndex;
    a->d = num;
    Print();
    a->l = DFS(left, left + lNum -1, numIndex - lNum, numIndex-1);
    a->r = DFS(right-rNum, right-1, numIndex+1, numIndex+rNum);
    return a;
}

int main(){
//  freopen("test.txt", "r", stdin);
    ios::sync_with_stdio(false);
    
    cin>>n;
    for(int i = 0; i < n; i++){
        cin>>back[i];
    }
    for(int i = 0; i < n; i++){
        cin>>mid[i];
    }
    root = new Node();
    root = DFS(0, n-1, 0, n-1);
    Print();
    return 0;
}

exprience

• 二叉樹要多寫幾回，就熟悉了。