Problem of the day
You have been given a Binary Tree of distinct integers and two nodes ‘X’ and ‘Y’. You are supposed to return the LCA (Lowest Common Ancestor) of ‘X’ and ‘Y’.
You may assume that given ‘X’ and ‘Y’ definitely exist in the given binary tree.
For example :
For the given binary tree
LCA of ‘X’ and ‘Y’ is highlighted in yellow colour.
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 contains two integers ‘X’ and ‘Y’ denoting the two nodes of the binary tree.
For example, the input for the tree depicted in the below image would be :
1
2 3
4 -1 5 6
-1 7 -1 -1 -1 -1
-1 -1
Explanation :
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 first not-null node(of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.
The input ends when all nodes at the last level are null(-1).
Note :
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
Output Format :
Print an integer denoting the LCA of the given binary tree nodes.
Note :
You do not need to print anything, it has already been taken care of. Just implement the given function.
1 2 3 4 7 -1 -1 -1 -1 -1 -1
4 7
2
1 2 3 -1 -1 -1 -1
2 3
1
0 <= 'N' <= 3000
0 <= 'DATA' <= 10^4
Where 'DATA' is a node value of the binary tree.
Time Limit: 1 sec
Can you think of considering all the common ancestors and choosing the farthest one from the root node?
The basic idea of this approach is to list down all the ancestors of the given nodes. And then, we will choose the common ancestor from the two lists which is farthest from the root of the tree.
Consider the binary tree as shown in the above figure, where we are trying to find the LCA of X and Y. Let us try to find the path from the root node to X and Y respectively and store the nodes present in the path in two separate lists pathToX and pathToY. Observe that the first few nodes(here nodes in yellow colour) in the lists will be the same which are common ancestors of node X and Y. Now, we need a common ancestor which is farthest from the root node. It is quite clear from the above image that the last common node in the lists will be the LCA.
Consider the following steps to describe the approach explained above:
O(N), Where N is the number of nodes in the given binary tree.
Since we are doing two pre-order traversals and in the worst case (Skewed Trees), all nodes of the given tree can be present in the path when ‘X’ and ‘Y’ are very close to the leaf node of the binary tree. And each one can take O(N) time. So the overall time complexity will be O(N).
O(N), Where N is the number of nodes in the given binary tree.
Since we are storing the paths in a list and in the worst case (Skewed Trees), all nodes of the given tree can be present in the list when ‘X’ will be the root node and ‘Y’ will be the leaf node of the binary tree. So the overall space complexity will be O(N).