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.
Related
There are preorder, inorder and postorder traversal for a binary tree, but no matter what order, it just traverses the tree to find a matched path. Is there any use case where I have to use any of the orders? Or are they just different ways but no difference regarding practical usage? Thanks.
There is definite practical usage with these traversals.
There are few specific use cases as below :
By using In-order traversal, you can get sorted node values if your requirement needs sorted information..
By using Pre-order traversal, you can create a copy of the tree and also can be used to get prefix expression of an expression tree.
Postorder traversal is used to delete the tree and also can be useful to get postfix expression of an expression tree.
The appropriate traversal technique shall be used based on which nodes should be fetched first for the requirement / design in hand. In case, if your requirement requires roots to be processed /picked / analyzed before picking up leaf nodes then pre-order traversal shall be helpful. Else, if leaf nodes have to be processed / fetched / analyzed before root nodes, then post-order shall be helpful.
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 am reading cracking the coding interview, and I have a question on the solution of the following problem:
You have two very large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of T1.
The simple solution that it suggests is to create a string representing the in-order and pre-order traversals and check if T2s pre-order/in-order traversal is substring of T1's pre-order/in-order traversal.
What I wonder is why do we need to compare both traversals? And why exactly that two traversals, why not for example in-order and post-order. And mainly won't only one traversal be enough? Say only in-order or pre-order traversal?
One traversal isn't enough. Consider the graphs 1->2->3 and 2<-1->3. If you start with node 1 and do a traversal, you encounter the nodes in the order 1, 2, 3. If you simply create a string showing the pre-order the two give the same result: 1,2,3
On the other hand, if you use a post-order, then the two will give a different result. 3,2,1 and 2,3,1
I bet for any one ordering, you can find two different trees with the same result.
So the question you need to answer for yourself for any other pair you want to look at is: would there be a tree that would give the same order for both traversals? I'm going to leave that as something to think about and come back later to see if you've got it.
I think preorder traversal with sentinel to represent null node is enough.
we can use this approach to serialize/deserialize a binary tree. That means, it is an one-to-one mapping between a binary tree to its preorder+sentinel representation.
After we get strings for both small tree and big tree. then we do a string match using kmp algorithm.
I know people are saying that we have to use both preorder and inorder (or postorder and inorder). but most of them just follow what others are saying, rather than think independently.
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.
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.