Bottom View of Binary Tree

 https://workat.tech/problem-solving/practice/bottom-view-binary-tree


There are different ways to look at a binary tree. The bottom view of a binary tree contains the set of nodes that will be visible if you look at the binary tree from the bottom.

Note: If there are multiple bottom-most nodes for a horizontal distance from root, use the later one in level-order traversal.

Given the root node of a binary tree, return an array containing the node elements in the bottom view, from left to right.

bottom-view-binary-tree

Testing

Input Format

The first line contains an integer T denoting the number of test cases.

For each test case, the input has 2 lines:

  • The first line contains an integer n denoting the number of nodes in the tree (including the NULL nodes).
  • The second line contains n space-separated integers that will form the binary tree. The integers follow level order traversal of the tree where -1 indicates a NULL node.

Output Format

For each test case, the output contains a line with space-separated integers representing the bottom view of the binary tree.

Sample Input

5
7
1 2 -1 4 -1 5 6
3
6 -1 4
7
8 -1 9 -1 10 11 12
5
28 14 11 -1 48
1
6

Expected Output

5 4 6 1
6 4
8 11 10 12
14 48 11
6

1 <= T <= 10
1 <= n <= 105
1 <= node value <= 106

---
Intuition
Level Order  Q based BFS
Pair (Node, col)

Map<Integer, Integer> - col to node.val

At each col, the last node.val is visible from bottom
--
Time Complexity - O(N)
Space Complexity - O(N)
--

Related problems
class Solution {
/* This is the Node class definition
class Node {
public Node left;
public Node right;
public int data;
public Node(int data) {
this.data = data;
}
}
*/
class Pair {
Node node;
int col;
Pair(Node n, int c) {
node = n;
col = c;
}
}
int[] bottomView(Node root) {
if (root == null) {
return new int[0];
}
List<Integer> list = new ArrayList<Integer>();
Queue<Pair> q = new LinkedList<Pair>();
q.offer(new Pair(root, 0));
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
int min = Integer.MAX_VALUE, max = Integer.MIN_VALUE;
while (!q.isEmpty()) {
int size = q.size();
for (int i = 0; i < q.size(); i++) {
Pair pair = q.poll();
min = Math.min(min, pair.col);
max = Math.max(max, pair.col);
map.put(pair.col, pair.node.data);
if (pair.node.left != null) {
q.offer(new Pair(pair.node.left, pair.col - 1));
}
if (pair.node.right != null) {
q.offer(new Pair(pair.node.right, pair.col + 1));
}
}
}
int[] ans = new int[map.size()];
for (int i = min, j = 0; i <= max; i++, j++) {
ans[j] = map.get(i);
}
return ans;
}
}