Problem of the day
1. A node will be in the bottom-view if it is the bottom-most node at its horizontal distance from the root.
2. The horizontal distance of the root from itself is 0. The horizontal distance of the right child of the root node is 1 and the horizontal distance of the left child of the root node is -1.
3. The horizontal distance of node 'n' from root = horizontal distance of its parent from root + 1, if node 'n' is the right child of its parent.
4. The horizontal distance of node 'n' from root = horizontal distance of its parent from the root - 1, if node 'n' is the left child of its parent.
5. If more than one node is at the same horizontal distance and is the bottom-most node for that horizontal distance, including the one which is more towards the right.
Input: Consider the given Binary Tree:
Output: 4 2 6 3 7
Explanation:
Below is the bottom view of the binary tree.
1 is the root node, so its horizontal distance = 0.
Since 2 lies to the left of 0, its horizontal distance = 0-1= -1
3 lies to the right of 0, its horizontal distance = 0+1 = 1
Similarly, horizontal distance of 4 = Horizontal distance of 2 - 1= -1-1=-2
Horizontal distance of 5 = Horizontal distance of 2 + 1= -1+1 = 0
Horizontal distance of 6 = 1-1 =0
Horizontal distance of 7 = 1+1 = 2
The bottom-most node at a horizontal distance of -2 is 4.
The bottom-most node at a horizontal distance of -1 is 2.
The bottom-most node at a horizontal distance of 0 is 5 and 6. However, 6 is more towards the right, so 6 is included.
The bottom-most node at a horizontal distance of 1 is 3.
The bottom-most node at a horizontal distance of 2 is 7.
Hence, the bottom view would be 4 2 6 3 7
The only line contains elements 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.
For example, the input for the tree depicted in the below image will 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).
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
Return an array representing the bottom view of the given binary tree.
You do not need to print anything; it has already been taken care of. Just implement the given function.
1 2 3 -1 -1 5 6 7 8 -1 -1 -1 -1 -1 -1
7 5 8 6
Test case 1:
As shown in the above figure,
1 is the root node, so its horizontal distance = 0.
Since 2 lies to the left of 0, its horizontal distance = 0-1= -1
3 lies to the right of 0, its horizontal distance = 0+1 = 1
Similarly, horizontal distance of 5 = Horizontal distance of 3 - 1= 1-1= 0
Horizontal distance of 6 = Horizontal distance of 3 + 1= 1+1 = 2
Horizontal distance of 7 = 0-1 =-1
Horizontal distance of 8 = 0+1 = 1
The bottom-most node at a horizontal distance of -1 is 7.
The bottom-most node at a horizontal distance of 0 is 5.
The bottom-most node at a horizontal distance of 1 is 8.
The bottom-most node at a horizontal distance of 2 is 6.
Hence, the bottom view would be 7 5 8 6.
1 2 3 4 -1 6 7 -1 -1 -1 8 -1 -1 -1 -1
4 2 6 8 7
Try to do this in O(n*log(n)).
1 <= n <= 10000
Where 'n' is the total number of nodes in the binary tree.
Time Limit: 1 sec