


For the given binary tree: [1, 2, 3, -1, -1, 4, 5, -1, -1, -1, -1]
Start Node: 3
1
/ \
2 3
/ \
4 5
Output: 2
Explanation :
In the zeroth minute, Node 3 will start to burn.
After one minute, Nodes (1, 4, 5) that are adjacent to 3 will burn completely.
After two minutes, the only remaining Node 2 will be burnt and there will be no nodes remaining in the binary tree.
So, the whole tree will burn in 2 minutes.
The first line contains elements of the tree in the level order form. The line consists of values of nodes separated by a single space. In case a node is null, we take -1 in its place.
The second line of input contains the value of the start node.

For example, the input for the tree depicted in the above image would be :
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Level 1 :
The root node of the tree is 1
Level 2 :
Left child of 1 = 2
Right child of 1 = 3
Level 3 :
Left child of 2 = 4
Right child of 2 = null (-1)
Left child of 3 = 5
Right child of 3 = 6
Level 4 :
Left child of 4 = null (-1)
Right child of 4 = 7
Left child of 5 = null (-1)
Right child of 5 = null (-1)
Left child of 6 = null (-1)
Right child of 6 = null (-1)
Level 5 :
Left child of 7 = null (-1)
Right child of 7 = null (-1)
The above format was just to provide clarity on how the input is formed for a given tree.
The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:
1 2 3 4 -1 5 6 -1 7 -1 -1 -1 -1 -1 -1
Print a single integer denoting the time in minutes that will be taken to burn the whole tree.
You do not need to print anything, it has already been taken care of. Just implement the given function.
The idea is to first create an undirected graph of the given binary tree and then doing a bfs traversal of the undirected graph starting from the start node. We will keep a variable ‘count’ that will be incremented at each level of bfs traversal. ‘count-1’ is the required time needed to burn the whole tree.
Algorithm
The total time taken by the tree to burn completely will be equal to the distance of the farthest node from the start node in the tree. So we will find three values for each node.
Algorithm
For the given binary tree and Start node 6
1 2
/ \ Value of Above for each Node: / \
2 3 -1 1
/ \ / \ / \ / \
4 5 6 7 -1 -1 0 -1
/ \ / \
8 9 -1 -1
For the given binary tree and Start Node 3,
1 2
/ \ Value of Below for each Node: / \
2 3 1 2
/ \ / \ / \ / \
4 5 6 7 0 0 1 1
/ \ / \
8 9 0 0
Two Sum IV - Input is a BST
Icarus and BSTCOUNT
Height of Binary Tree
Height of Binary Tree
Height of Binary Tree
Height of Binary Tree
Locked Binary Tree
Maximum Island Size in a Binary Tree