I'm looking at an interview book and the question is:
You have two very large binary trees:
T1
, with millions of nodes, andT2
, with hundreds of nodes. Create an algorithm to decide ifT2
is a subtree ofT1
.
The authors mentions this as a possible solution:
Note that the problem here specifies that
T1
has millions of nodes—this means that we should be careful of how much space we use. Let’s say, for example,T1
has 10 million nodes—this means that the data alone is about40 mb
. We could create a string representing the inorder and preorder traversals. IfT2
’s preorder traversal is a substring ofT1
’s preorder traversal, andT2
’s inorder traversal is a substring ofT1
’s inorder traversal, thenT2
is a substring ofT1
.
I'm not quite sure the logic behind as to why if these are true:
T2-preorder-traversal-string
is a substring of T1-preorder-traversal-string
T2-inorder-traversal-string
is a substring of T1-inorder-traversal-string
That T2
must be a substring (although I assume the author means subtree) of T1
. Can I get an explanation to this logic?
EDIT: User BartoszMarcinkowski brings up a good point. Assume both trees have no duplicate nodes.
In-order traversal is very commonly used in binary search trees because it returns values from the underlying set in order, according to the comparator that set up the binary search tree (hence the name). Post-order traversal while deleting or freeing nodes and values can delete or free an entire binary tree.
For Inorder, you traverse from the left subtree to the root then to the right subtree. For Preorder, you traverse from the root to the left subtree then to the right subtree. For Post order, you traverse from the left subtree to the right subtree then to the root.
Preorder traversal is used to create a copy of the tree. Preorder traversal is also used to get prefix expression on an expression tree.
What is the speciality about the inorder traversal of a binary search tree? Explanation: As a binary search tree consists of elements lesser than the node to the left and the ones greater than the node to the right, an inorder traversal will give the elements in an increasing order.
I think it is not true. Consider:
T2:
2
/ \
1 3
inorder 123 preorder 213
and
T1:
0
/ \
3 3
/ \
1 1
/ \
0 2
inorder 0123103 preorder 0310213
123
is substring of 0123103
, 213
is substring of 0310213
, but T2 is not subtree of T1.
Here is a counter-example to the method.
Consider the tree T1
:
B
/ \
A D
/ \
C E
\
F
And the sub-tree T2
:
D
/ \
C E
The relevant traversals are:
T1
pre-order: BADCEF
T2
pre-order: DCE
T1
in-order: ABCDEF
T2
in-order: CDE
While DCE
is in BADCEF
and CDE
is in ABCDEF
, T2
is not actually a sub-tree of T1
. The author's definition of sub-tree must have been different or it was just a mistake.
Related question: Determine if a binary tree is subtree of another binary tree using pre-order and in-order strings
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With