A book I'm reading claims that one way to check whether a binary tree B is a subtree of a binary tree A is to build the inorder and preorder strings (strings that represent the inorder and preorder traversal of each tree) of both trees, and check whether inorder_B is a substring of inorder_A and preorder_B is a substring of preorder_A. Note that it claims that you have to check substring match on both the inorder and preorder strings.
Is it really necessary to check for a substring match on both the inorder and preorder strings? Wouldn't it suffice to check either? Could someone provide an example to prove me wrong (i.e. prove the claim in the book right)? I couldn't come up with an example where two trees were unequal but the preorder or inorder strings match.
Consider the two two node trees with A and B as nodes. Tree one has B as root and A as left child. Tree two has A as root and B as right child. The inorder traversals match but the trees differ.
I think you need both if the tree is not a binary search tree but a plain binary tree. Any set of nodes can be a preorder notation. Suppose there is a binary tree a,b,c,d,e,f,g,h and your subtree is cdef. You can create another tree with subtree cde and another subtree fg. There is no way to know the difference.
If it is a binary search tree however you needn't have the inorder.
Incidentally here's an amusing algorithmic problem: given a preorder notation find the number of binary trees that satisfy it.
As supplement to user1952500's answer: if it is a binary search tree, either only preorder or only postorder can make it unique, while only inorder can not. For example:
5
/ \
3 6
inorder: 3-5-6
however, another binary search tree can have the same inorder:
3
\
5
\
6
Also, I believe preorder+inorder+string_comparison only works to check whether two trees are identical. It fails to check whether a tree is the subtree of another tree. To see an example, refer
Determine if a binary tree is subtree of another binary tree using pre-order and in-order strings
a preorder traversal with sentinel to represent null node is well enough.
we can use this approach to serialize/deserialize binary tree. it means there is one-to-one mapping between a binary tree and its preorder with sentinel representation.
Related
The below is an excerpt from geeksforgeeks
If the given Binary Tree is Binary Search Tree, we can store it by
either storing preorder or postorder traversal. In case of Binary
Search Trees, only preorder or postorder traversal is sufficient to
store structure information.
Questions
Is it not possible to use in-order traversal for serialization and deserialization of Binary Trees?. if so why?
what is the distinction between Binary Tree and BST serialization?. the above statement is not clear about this distinction
In-order traversal of a BST produces the sorted list of data, regardless of how the tree looked like.
On the contrary, given a list produced by pre-order traversal, the BST can be reconstructed:
The first element is a root.
Split the rest by the value of the root. The S of BST guarantees that the split point exists, and the first/second slices encompass the left/right subtrees respectively.
Recursively apply the procedure to first and second slice.
This procedure relies heavily on the S property of BST. An arbitrary Binary Tree doesn't have the split point.
Assume you have two binary trees and you want to know whether one is a subtree of the other. One solution is to get the inorder and preorder traversals of both trees and check whether the traversals of the candidate subtree are substrings of the corresponding traversal for the other tree. I read several posts about this posts about this solution. One discussion shows that inorder AND preorder traversal are both necessary. Can someone explain why they are sufficient? Why is the case that if the inorder and preorder traversal of tree2 are substrings of those of tree1, then tree2 is a subtree of tree1?
Q: One discussion shows that inorder AND preorder traversal are both
necessary. Can someone explain why they are sufficient?
Because of the simple fact that it is possible to uniquely reconstruct a binary tree from these two traversals (or inorder and postorder, as well). Check this example:
Inorder : [1,2,3,4,5,6]
Preorder : [4,2,1,3,5,6]
From preorder, you know that 4 is the root of the tree. From inorder, you can determine the left and right subtree, and you proceed recursively from this point:
4
/ \
Left subtree Right subtree
Inorder : [1,2,3] Inorder : [5,6]
Preorder: [2,1,3] Preorder: [5,6]
Check for more details in this excellent article:
Reconstructing binary trees from tree traversal. Since these two serializations (traversals actually serialize tree to a string) of the tree combined together have to be unique for a binary tree, we get that one tree is a subtree of another if and only if these traversals are substrings of other two serializations.
People agreed that binary tree can represent the order on it's nodes by left/right relation. That means that left part comes before right part. You may call trees equivalent if the order is the same. So in-order string represents the the order and if you want to check the equivalence, then it is sufficient to check only in-order (by definition).
But when you want to check the full equality of trees then we have to find the way how we can distinguish equivalent trees.For example it can be level-order check. But for subtrees level order doesn't fit, because the level order string for subtree is split. For pre-order you walk the subtree form root before other parts of tree.
Suppose equivalent trees are not equal, then traversing in pre-order everything will be equal until first differ. 2 situations can happen.
1) The value of node of one tree differs from another. That means that pre-order strings differs, because you walk the tree in pre-order.
2) Children signature (no children, only left, only right, both children) differs. But in this situation easy to understand that the in-order will change and trees are not equivalent, which contradicts the conditions.
Note that this works only when all the nodes are unique. If you have all nodes of value like "a" then no matter how you walk, your string is always "aa...a". So you have to distinguish the nodes somehow, not only by "value".
I have Binary Search tree and have to perform three types of tree traversal:
Are this results correct?
Pre-order (root,left,right): 30,15,59,43,40,92
In-order (left,root,right): 15,30,59,40,43,92
Post-order (left,right,root): 15,59,40,43,92,30
UPDATE:
In-order: 15,30,40,43,59,92 (projection?)
Post-order: 15,40,43,92,59,30.
Is it right?
Given this updated tree, your preorder traversal is correct.
Your inorder traversal, though, is incorrect. As a hint, doing an inorder traversal of a binary tree always lists the values off in sorted order.
Finally, your postorder traversal is incorrect. The value 59 won't be produced until after all of the nodes in its two subtrees are produced, so it should come second-to-last. Using this fact, try seeing if you can come up with the correct answer.
Hope this helps!
I am trying to get straight in my head how tree traversals can be used to uniquely identify a tree, and the crux of it seems to be whether the tree is a vanilla Binary Tree (BT), or if it also has the stricter stipulation of being a Binary Search Tree (BST). This article seems to indicate that for BT's, a single inorder, preorder and postorder traversal will not uniquely identify a tree (uniquely means structure and values of keys in this context). Here is a quick summary of the article:
BTs
1. We can uniquely reconstruct a BT with preorder + inorder and postorder + inorder.
2. We can also use preorder + postorder if we also stipulate that the traversals keeps track of the null children of a node.
(an open question (for me) is if the above is still true if the BT can have non-unique elements)
BSTs
3. We cannot use inorder for a unique id. We need inorder + preorder, or inorder + postorder.
Now, (finally) my question is, can we use just pre-order or just post-order to uniquely identify a BST? I think that we can, since this question and
answer
seems to say yes, we can use preorder, but any input much appreciated.
I can't tell what's being asked here. Any binary tree, whether it's ordered or not, can be serialized by writing out a sequence of operations needed to reconstruct the tree. Imagine a simple stack machine with just two instructions:
Push an empty tree (or NULL pointer if you like) onto the stack
Allocate a new internal node N, stuff a value into N, pop the top two trees off the stack and make them N's left and right children, and finally push N onto the stack.
Any binary tree can be serialized as a "program" for such a machine.
The serialization algorithm uses a postorder traversal.
Okay, you can use preorder only to identify a tree. This is possible because only in preorder traversal does the id-of-current-node comes before the ids of children. So you can read the traversal output root-to-leaves.
You can check http://en.wikipedia.org/wiki/Tree_traversal#Pre-order to confirm
So you can consider a preorder traversal as a list of insertions into a tree. Because the tree insertion into BST is deterministic, when you insert a list of values into an empty tree, you always get the same tree.
In-order tree traversal obviously has application; getting the contents in order.
Preorder traversal seems really useful for creating a copy of the tree.
Is there a common use for postorder traversal of a binary tree?
Let me add another one:
Postorder traversal is also useful in deleting a tree. In order to free up allocated memory of all nodes in a tree, the nodes must be deleted in the order where the current node can only be deleted when both of its left and right subtrees are deleted.
Postorder does exactly just that. It processes both of the left and right subtrees before processing the current node.
If the tree represents a mathematical expression, then to evaluate the expression, a post-order traversal is necessary.
Yes. Postorder is sometimes used to translate mathematical expressions between different notations.
It can also generate a postfix representation of a binary tree.