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
给出后序与中序遍历,要求层次遍历。
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
- 二叉树要多写几次,就熟悉了。
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