The following is from Max Howell @twitter:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can't invert a binary tree on a whiteboard so fuck off.
Now it's your turn to prove that YOU CAN invert a binary tree!
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) which is the total number of nodes in the tree -- and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node from 0 to N-1, and gives the indices of the left and right children of the node. If the child does not exist, a "-" will be put at the position. Any pair of children are separated by a space.
Output Specification:
For each test case, print in the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.
Sample Input:
8
1 -
- -
0 -
2 7
- -
- -
5 -
4 6
Sample Output:
3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1
#include<iostream>
#include<list>
using namespace std;
struct Node{
int left;
int right;
};
typedef Node* PNode;
int root;
int N;
int flag = false;
void levelOrder(PNode tree)
{
int r;
list<int> q;
q.push_back(root);
while(!q.empty())
{
if(flag == true)
cout << " ";
else
flag = true;
r = q.front();
q.pop_front();
cout << r;
if(tree[r].left != -1)
q.push_back(tree[r].left);
if(tree[r].right != -1)
q.push_back(tree[r].right);
}
flag = false;
cout << endl;
}
int toNum(char c)
{
if(c=='-') return -1;
return (c - '0');
}
void inorder(int r,PNode tree)
{
if(tree[r].left != -1)
inorder(tree[r].left,tree);
if(flag != false)
cout << " ";
else
flag = true;
cout << r;
if(tree[r].right != -1)
inorder(tree[r].right,tree);
}
int main()
{
int l,r;
bool * findRoot;
char c1,c2,c3;
cin >> N;
findRoot = new bool[N];
PNode tree = new Node[N];
for(int i=0; i<N; i++)
{
cin >> c1 >> c2;
l = toNum(c1);
r = toNum(c2);
if(l!=-1) findRoot[l] = true;
if(r!=-1) findRoot[r] = true;
tree[i].left = r;
tree[i].right = l;
}
for(int i=0; i<N; i++)
if(findRoot[i] == false)
{
root = i;
break;
}
levelOrder(tree);
inorder(root,tree);
return 0;
}
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