Given a binary tree and a leaf node from this tree. It is known that in 1s all nodes connected to a given node (left child, right child and parent) get burned in 1 second. Then all the nodes which are connected through one intermediate get burned in 2 seconds, and so on. The task is to find the minimum time required to burn the complete binary tree.
Input : 1 / \ 2 3 / \ \ 4 5 6 / \ \ 7 8 9 \ 10 Leaf = 8 Output : 7 Initially 8 is set to fire at 0th sec. 1 / \ 2 3 / \ \ 4 5 6 / \ \ 7 F 9 \ 10 After 1s: 5 is set to fire. 1 / \ 2 3 / \ \ 4 F 6 / \ \ 7 F 9 \ 10 After 2s: 2, 7 are set to fire. 1 / \ F 3 / \ \ 4 F 6 / \ \ F F 9 \ 10 After 3s: 4, 1 are set to fire. F / \ F 3 / \ \ F F 6 / \ \ F F 9 \ 10 After 4s: 3 is set to fire. F / \ F F / \ \ F F 6 / \ \ F F 9 \ 10 After 5s: 6 is set to fire. F / \ F F / \ \ F F F / \ \ F F 9 \ 10 After 6s: 9 is set to fire. F / \ F F / \ \ F F F / \ \ F F F \ 10 After 7s: 10 is set to fire. F / \ F F / \ \ F F F / \ \ F F F \ F It takes 7s to burn the complete tree.
The idea is to store additional information for every node:
- Depth of left subtree.
- Depth of right subtree.
- The time required for the fire to reach the current node starting from the first leaf node burned.
- A boolean variable to check if the initial burnt node is in the tree rooted under current node.
Before moving ahead with the approach let’s take a look at the tree below:
1 / \ 2 3 / \ / 4 5 6 / / \ 8 9 10 / 11
In the above tree, if we set the leaf node 11 at fire.
- In 1s, the fire will reach node 9.
- In 2s, the fire will reach node 5.
- In 3rd second, the fire will reach node 2 and 10. Here comes an observation:
- In 2s fire reached node 5. For node 5, the initial burned leaf is in it’s left subtree, so the time taken to burn right subtree will be the height of the right subtree which is 1. Therefore, fire reaches to node 10 in (2+1) = 3s.
- Again, for the node 2. Fire reached to node 2 in 3s from right subtree. Therefore, time taken to burn left subtree will be it’s height.
So the solution is to apply recursion and for every node calculate the below-required values:
- Left Depth.
- Right Depth.
- The time required for fire to reach the current node.
- Is the current subtree conatins initial burnt leaf node.
So, for the minimum time required to burn any subtree will be:
The time required for fire to reach the root node from initial burnt leaf + depth of the opposite side
Therefore, to find time required to burn the complete tree, we need to calculate the above value for every node, and take maximum of that value.
ans = max(ans, (time required for fire to reach current node + depth of other subtree))
Below is the implementation of the above approach:
- Burn the binary tree starting from the target node
- Queries to find sum of distance of a given node to every leaf node in a Weighted Tree
- Get last node of the binary tree following given pattern starting from X
- Delete the last leaf node in a Binary Tree
- Closest leaf to a given node in Binary Tree
- Deepest left leaf node in a binary tree
- Print the nodes of binary tree as they become the leaf node
- Number of leaf nodes in the subtree of every node of an n-ary tree
- Deepest right leaf node in a binary tree | Iterative approach
- Deepest left leaf node in a binary tree | iterative approach
- Create a binary tree from post order traversal and leaf node array
- Generate Complete Binary Tree in such a way that sum of non-leaf nodes is minimum
- Print the longest leaf to leaf path in a Binary tree
- Minimum time required to visit all the special nodes of a Tree
- Check whether a node is leaf node or not for multiple queries
- Find the node with minimum value in a Binary Search Tree
- Minimum valued node having maximum depth in an N-ary Tree
- Find the node with minimum value in a Binary Search Tree using recursion
- Construct XOR tree by Given leaf nodes of Perfect Binary Tree
- Minimum and maximum node that lies in the path connecting two nodes in a Binary Tree
Improved By : princiraj1992