Problem #
Source: [PAT 1102]
Description #
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 6Sample Output: #
3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1Solution #
Code #
#include <string>
#include <iostream>
#include <vector>
#include <queue>
#define max_size 11
using namespace std;
bool isRoot[max_size];
int n, index;
struct node
{
int v, left, right;
node()
{
v = -1;
left = -1;
right = -1;
}
};
node tree[max_size];
void postOrder(node &root)
{
if (root.v == -1)
{
return;
}
if (root.left != -1)
postOrder(tree[root.left]);
if (root.right != -1)
postOrder(tree[root.right]);
int tmp = root.right;
root.right = root.left;
root.left = tmp;
return;
}
void level(int root)
{
queue<int> q;
q.push(root);
while (!q.empty())
{
int now = q.front();
q.pop();
if (tree[now].left != -1)
q.push(tree[now].left);
if (tree[now].right != -1)
q.push(tree[now].right);
cout<<tree[now].v;
if (q.size() != 0)
{
cout << " ";
}
}
}
void inOrder(node root)
{
if (root.v == -1)
{
return;
}
if (root.left != -1)
inOrder(tree[root.left]);
cout << root.v;
if (index < n)
cout << " ";
++index;
if (root.right != -1)
inOrder(tree[root.right]);
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n;
string s;
fill(isRoot, isRoot + n, true);
for (int i = 0; i < n; i++)
{
cin >> s;
tree[i].v = i;
if (s[0] != '-')
{
int item = s[0] - '0';
isRoot[item] = false;
tree[i].left = item;
}
cin >> s;
if (s[0] != '-')
{
int item = s[0] - '0';
isRoot[item] = false;
tree[i].right = item;
}
}
int root;
for (int i = 0; i < n; i++)
{
if (isRoot[i])
{
root = i;
break;
}
}
postOrder(tree[root]);
index = 1;
level(root);
cout<<endl;
inOrder(tree[root]);
return 0;
}
