Print the Bottom View of the Binary Tree.

Objec­tive: Given a binary tree, print it in Bottom View of it.

What is Bottom View: Bottom view means when you look the tree from the bottom the nodes you will see will be called the bottom view of the tree. See the exam­ple below.

Bottom View of a Binary Treeas you can see in the example above,8, 4, 9, 5, 3, 7 is the bottom view of the given binary tree.

Approach:

This Approach is quite sim­i­lar to the Print the Binary Tree in Ver­ti­cal Order Path.
and Print The Top View of a Binary Tree. Just mod­i­fied the code so that it will print only the last ele­ment it will encounter in the ver­ti­cal order. That will be the bottom view.

How will you know that you are vis­it­ing the last node at every level(Vertically)?

  • Take a variable called level, whenever you go left, do level++ AND whenever you go right do level–.
  • With step above we have separated out the levels vertically.
  • level distribution

    level distribution

  • Now you need to store the ele­ments of each level, so cre­ate a TreeMap and the (key,value) pair will be (level,element at that level).
  • Now all we need to do the level order tra­ver­sal and store only recent visited node at each level(vertically), this way you will be storing only the last element at each level.
  • We will use sim­ple queue tech­nique for level order tra­ver­sal or BFS.
  • we will cre­ate a class QueuePack, this will store the objects con­tain­ing node and its level.
  • At the end traverse through TreeMap and print all the values in it, it will be the bottom view of a tree.
  • See the code for bet­ter understanding.

Code:

Output:
4 6 5 7 8 

 

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If you find anything incorrect or you feel that there is any better approach to solve the above problem, please write comment.
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  • Xiaozhen Zhong

    how would the tpre-order traverse guarantee most recent visited node would be at “bottom” of a vertical level?
    Would following sequence work?
    Node root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.left.right = new Node(5);
    root.left.right.left = new Node(6);
    root.left.right.right = new Node(7);
    root.left.right.right.right = new Node(9); ==> add another right node, which will be at same vertical level of node(8) but lower.
    root.right.right = new Node(8);
    We should get 4 6 5 7 9 as bottom view, but my guess is the code will return 4 6 5 7 8 (wrong result)

    • tutorialhorizon

      Hi Xiaozhen,

      Code is correct and will return 4 6 5 7 9, don’t concentrate on preorder traversal, we are storing the node at each vertical level in a hash map ( in fact we are updating it so hash map will contain the last node visited at that level) See the diagram.

  • http://www.waytocrack.com kumar
  • Subodh Karwa

    Can you please explain how 5 and 6 both being at level -0 are printed using the code ?

  • suvro

    Print In-Order nodes where either left or right child is NULL

    private static void printBottomView(Tree tree) {

    if(tree==null)
    return;
    printBottomView(tree.leftChild);
    if(tree.leftChild==null||tree.rightChild==null)
    System.out.println(“===”+tree.data+”===”);
    printBottomView(tree.rightChild);
    }

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