Evaluating an expression tree without a stack - algorithm

Is it possible to evaluate an expression tree (pre/postfix) without using a stack? Had this question while talking about trees in the algorithm class at school. My guess is no.

Yes, you can.
Do a breadth first traversal of the tree (like a search, but do through all the tree). You can do this with a vector/queue/list in a iterative way.
Once you are done you can go backwards through the list/vector/queue you generated in the previous step. At each point compute the value of the node in the list. Since you have all the children already visited (you are going backwards) all you have to do is to lookup their value and apply the instructions in the node.

Related

Mirror a binary tree in constant time

This is not a homework question. I heard that it is possible to mirror a binary tree i.e. flip it, in constant time. Is this really the case?
Sure, depending on your data structure, you would just do the equivalent of: instead of traversing down the left node and then the right node, you would traverse down the right node, and then the left node. This could be a parameter passed into the recursive function that traverses the tree (i.e. in C/C++, a bool bDoLeftFirst, and an if-statement that uses that parameter to decide which order to traverse the child nodes in).
Did you mean "invert binary tree", the problem which Max Howell could not solve and thus rejected by Google?
https://leetcode.com/problems/invert-binary-tree/
You can find solutions in the "discuss" section.

Threaded Binary Search Trees Advantage

An explanation about Threaded Binary Search Trees (skip it if you know them):
We know that in a binary search tree with n nodes, there are n+1 left and right pointers that contain null. In order to use that memory that contain null, we change the binary tree as follows -
for every node z in the tree:
if left[z] = NULL, we put in left[z] the value of tree-predecessor(z) (i.e, a pointer to the node which contains the predecessor key),
if right[z] = NULL, we put in right[z] the value of tree-successor(z) (again, this is a pointer to the node which contains the successor key).
A tree like that is called a threaded binary search tree, and the new links are called threads.
And my question is:
What is the main advatage of Threaded Binary Search Trees (in comparison to "Regular" binary search trees).
A quick search in the web has told me that it helps to implement in-order traversal iteratively, and not recursively.
Is that the only difference? Is there another way we can use the threads?
Is that so meaningful advantage? and if so, why?
Recursive traversal costs O(n) time too, so..
Thank you very much.
Non-recursive in-order scan is a huge advantage. Imagine that somebody asks you to find the value "5" and the four values that follow it. That's difficult using recursion. But if you have a threaded tree then it's easy: do the recursive in-order search to find the value "5", and then follow the threaded links to get the next four values.
Similarly, what if you want the four values that precede a particular value? That's difficult with a recursive traversal, but trivial if you find the item and then walk the threaded links backwards.
The main advantage of Threaded Binary Search Trees over Regular one is in Traversing nature which is more efficient in case of first one as compared to other one.
Recursively traversing means you don't need to implement it with stack or queue .Each node will have pointer which will give inorder successor and predecessor in more efficient way , while implementing traversing in normal BST need stack which is memory exhaustive (as here programming language have to consider implementation of stack) .

Fair deletion of nodes in Binary Search Tree

The idea of deleting a node in BST is:
If the node has no child, delete it and update the parent's pointer to this node as null
If the node has one child, replace the node with its children by updating the node's parent's pointer to its child
If the node has two children, find the predecessor of the node and replace it with its predecessor, also update the predecessor's parent's pointer by pointing it to its only child (which only can be a left child)
the last case can also be done with use of a successor instead of predecessor!
It's said that if we use predecessor in some cases and successor in some other cases (giving them equal priority) we can have better empirical performance ,
Now the question is , how is it done ? based on what strategy? and how does it affect the performance ? (I guess by performance they mean time complexity)
What I think is that we have to choose predecessor or successor to have a more balanced tree ! but I don't know how to choose which one to use !
One solution is to randomly choose one of them (fair randomness) but isn't better to have the strategy based on the tree structure ? but the question is WHEN to choose WHICH ?
The thing is that is fundamental problem - to find correct removal algorithm for BST. For 50 years people were trying to solve it (just like in-place merge) and they didn't find anything better then just usual algorithm (with predecessor/successor removing). So, what is wrong with classic algorithm? Actually, this removing unbalances the tree. After several random operations add/remove you'll get unbalanced tree with height sqrt(n). And it is no matter what you choosed - remove successor or predecessor (or random chose beetwen these ways) - the result is the same.
So, what to choose? I'm guessing random based (succ or pred) deletion will postpone unbalancing of your tree. But, if you want to have perfectly balanced tree - you have to use red-black ones or something like that.
As you said, it's a question of balance, so in general the method that disturbs the balance the least is preferable. You can hold some metrics to measure the level of balance (e.g., difference from maximal and minimal leaf height, average height etc.), but I'm not sure whether the overhead worth it. Also, there are self-balancing data structures (red-black, AVL trees etc.) that mitigate this problem by rebalancing after each deletion. If you want to use the basic BST, I suppose the best strategy without apriori knowledge of tree structure and the deletion sequence would be to toggle between the 2 methods for each deletion.

How to get parent in a k-ary level order succint trie?

I'm implementing a level order succint trie and I wan't to be able for a given node to jump back to his parent.
I tried several combination of rank/level but I can't wrap my head around this one...
I'm using this article as a base documentation :
http://stevehanov.ca/blog/index.php?id=120
It explain how to traverse childs, but not how to go up.
Thanks to this MIT lecture (http://www.youtube.com/watch?v=1MVVvNRMXoU) I know this is possible (in constant time as stated at 15:50), but the speaker only explain it for binary trie (eg: using the formula select1(floor(i/2)) ).
How can I do that on a k-ary trie?
Well, I don't know what select1() is, but the other part (floor(i/2)) looks like the trick you would use in an array-embedded binary tree, like those described here. You would divide by 2 because every parent has exactly 2 children --> every level uses twice the space of the parent level.
If you don't have the same number of children in every node (excepting leafs and perhaps one node with less children), you can't use this trick.
If you want to know the parent of any given node, you will need to add a pointer to the parent in every node.
Though, since trees are generally traversed starting at the root and going down, the usual thing to do is to store, in an array, the pointers to the nodes of the path. At any given point, the parent of the current node is the previous element in the array. This way you don't need to add a pointer to the parent in every node.
I think I've found my answer. This paper of Guy Jacobson explains it in section 3.2 Level-order unary degree sequence.
parent(x){ select1(rank0(x)) }
Space-efficient Static Trees and Graphs
http://www.cs.cmu.edu/afs/cs/project/aladdin/wwwlocal/compression/00063533.pdf
This work pretty good, as long as you don't mess up your node numbering like I was.

What's the difference between backtracking and depth first search?

What's the difference between backtracking and depth first search?
Backtracking is a more general purpose algorithm.
Depth-First search is a specific form of backtracking related to searching tree structures. From Wikipedia:
One starts at the root (selecting some node as the root in the graph case) and explores as far as possible along each branch before backtracking.
It uses backtracking as part of its means of working with a tree, but is limited to a tree structure.
Backtracking, though, can be used on any type of structure where portions of the domain can be eliminated - whether or not it is a logical tree. The Wiki example uses a chessboard and a specific problem - you can look at a specific move, and eliminate it, then backtrack to the next possible move, eliminate it, etc.
I think this answer to another related question offers more insights.
For me, the difference between backtracking and DFS is that backtracking handles an implicit tree and DFS deals with an explicit one. This seems trivial, but it means a lot. When the search space of a problem is visited by backtracking, the implicit tree gets traversed and pruned in the middle of it. Yet for DFS, the tree/graph it deals with is explicitly constructed and unacceptable cases have already been thrown, i.e. pruned, away before any search is done.
So, backtracking is DFS for implicit tree, while DFS is backtracking without pruning.
IMHO, most of the answers are either largely imprecise and/or without any reference to verify. So let me share a very clear explanation with a reference.
First, DFS is a general graph traversal (and search) algorithm. So it can be applied to any graph (or even forest). Tree is a special kind of Graph, so DFS works for tree as well. In essence, let’s stop saying it works only for a tree, or the likes.
Based on [1], Backtracking is a special kind of DFS used mainly for space (memory) saving. The distinction that I’m about to mention may seem confusing since in Graph algorithms of such kind we’re so used to having adjacency list representations and using iterative pattern for visiting all immediate neighbors (for tree it is the immediate children) of a node, we often ignore that a bad implementation of get_all_immediate_neighbors may cause a difference in memory uses of the underlying algorithm.
Further, if a graph node has branching factor b, and diameter h (for a tree this is the tree height), if we store all immediate neighbors at each steps of visiting a node, memory requirements will be big-O(bh). However, if we take only a single (immediate) neighbor at a time and expand it, then the memory complexity reduces to big-O(h). While the former kind of implementation is termed as DFS, the latter kind is called Backtracking.
Now you see, if you’re working with high level languages, most likely you’re actually using Backtracking in the guise of DFS. Moreover, keeping track of visited nodes for a very large problem set could be really memory intensive; calling for a careful design of get_all_immediate_neighbors (or algorithms that can handle revisiting a node without getting into infinite loops).
[1] Stuart Russell and Peter Norvig, Artificial Intelligence: A Modern Approach, 3rd Ed
According to Donald Knuth, it's the same.
Here is the link on his paper about Dancing Links algorithm, which is used to solve such "non-tree" problems as N-queens and Sudoku solver.
Backtracking, also called depth-first search
Backtracking is usually implemented as DFS plus search pruning. You traverse search space tree depth-first constructing partial solutions along the way. Brute-force DFS can construct all search outcomes, even the ones, that do not make sense practically. This can be also very inefficient to construct all solutions (n! or 2^n). So in reality as you do DFS, you need to also prune partial solutions, which do not make sense in context of the actual task, and focus on the partial solutions, which can lead to valid optimal solutions. This is the actual backtracking technique - you discard partial solutions as early as possible, make a step back and try to find local optimum again.
Nothing stops to traverse search space tree using BFS and execute backtracking strategy along the way, but it doesn't make sense in practice because you would need to store search state layer by layer in the queue, and tree width grows exponentially to the height, so we would waste a lot of space very quickly. That's why trees are usually traversed using DFS. In this case search state is stored in the stack (call stack or explicit structure) and it can't exceed tree height.
Usually, a depth-first-search is a way of iterating through an actual graph/tree structure looking for a value, whereas backtracking is iterating through a problem space looking for a solution. Backtracking is a more general algorithm that doesn't necessarily even relate to trees.
I would say, DFS is the special form of backtracking; backtracking is the general form of DFS.
If we extend DFS to general problems, we can call it backtracking.
If we use backtracking to tree/graph related problems, we can call it DFS.
They carry the same idea in algorithmic aspect.
DFS describes the way in which you want to explore or traverse a graph. It focuses on the concept of going as deep as possible given the choice.
Backtracking, while usually implemented via DFS, focuses more on the concept of pruning unpromising search subspaces as early as possible.
IMO, on any specific node of the backtracking, you try to depth firstly branching into each of its children, but before you branch into any of the child node, you need to "wipe out" previous child's state(this step is equivalent to back walk to the parent node). In other words, each siblings state should not impact each other. On the contrary, during normal DFS algorithm, you don't usually have this constraint, you don't need to wipe out(back track) previous siblings state in order to construct next sibling node.
Depth first is an algorithm for traversing or searching a tree. See here. Backtracking is a much more broad term that is used whereever a solution candidate is formed and later discarded by backtracking to a former state. See here. Depth first search uses backtracking to search a branch first (solution candidate) and if not successful search the other branch(es).
Depth first search(DFS) and backtracking are different searching and traversing algorithms. DFS is more broad and is used in both graph and tree data structure while DFS is limited to tree. With that being said,it does not mean DFS can't be used in graph. It is used in graph as well but only produce spanning tree, a tree without loop(multiple edge starting and ending at same vertex). That is why it is limited to tree.
Now coming back to backtracking, DFS use backtracking algorithm in tree data structure so in tree, DFS and backtracking are similar.
Thus,we can say, they are in same in tree data structure whereas in graph data structure, they are not same.
Idea - Start from any point, check if its the desired endpoint, if yes then we found a solution else goes to all next possible positions and if we can't go further then return to the previous position and look for other alternatives marking that current path wont lead us to the solution.
Now backtracking and DFS are 2 different names given to the same idea applied on 2 different abstract data types.
If the idea is applied on matrix data structure we call it backtracking.
If the same idea is applied on tree or graph then we call it DFS.
The cliche here is that a matrix could be converted to a graph and a graph could be transformed to a matrix. So, we actually apply the idea. If on a graph then we call it DFS and if on a matrix then we call it backtracking.
The idea in both of the algorithm is same.
The way I look at DFS vs. Backtracking is that backtracking is much more powerful. DFS helps me answer whether a node exists in a tree, while backtracking can help me answer all paths between 2 nodes.
Note the difference: DFS visits a node and marks it as visited since we are primarily searching, so seeing things once is sufficient. Backtracking visits a node multiple times as it's a course correction, hence the name backtracking.
Most backtracking problems involve:
def dfs(node, visited):
visited.add(node)
for child in node.children:
dfs(child, visited)
visited.remove(node) # this is the key difference that enables course correction and makes your dfs a backtracking recursion.
In a depth-first search, you start at the root of the tree and then explore as far along each branch, then you backtrack to each subsequent parent node and traverse it's children
Backtracking is a generalised term for starting at the end of a goal, and incrementally moving backwards, gradually building a solution.
Backtracking is just depth first search with specific termination conditions.
Consider walking through a maze where for each step you make a decision, that decision is a call to the call stack (which conducts your depth first search)... if you reach the end, you can return your path. However, if you reach a deadend, you want to return out of a certain decision, in essence returning out of a function on your call stack.
So when I think of backtracking I care about
State
Decisions
Base Cases (Termination Conditions)
I explain it in my video on backtracking here.
An analysis of backtracking code is below. In this backtracking code I want all of the combinations that will result in a certain sum or target. Therefore, I have 3 decisions which make calls to my call stack, at each decision I either can pick a number as part of my path to reach the target num, skip that number, or pick it and pick it again. And then if I reach a termination condition, my backtracking step is just to return. Returning is the backtracking step because it gets out of that call on the call stack.
class Solution:
"""
Approach: Backtracking
State
-candidates
-index
-target
Decisions
-pick one --> call func changing state: index + 1, target - candidates[index], path + [candidates[index]]
-pick one again --> call func changing state: index, target - candidates[index], path + [candidates[index]]
-skip one --> call func changing state: index + 1, target, path
Base Cases (Termination Conditions)
-if target == 0 and path not in ret
append path to ret
-if target < 0:
return # backtrack
"""
def combinationSum(self, candidates: List[int], target: int) -> List[List[int]]:
"""
#desc find all unique combos summing to target
#args
#arg1 candidates, list of ints
#arg2 target, an int
#ret ret, list of lists
"""
if not candidates or min(candidates) > target: return []
ret = []
self.dfs(candidates, 0, target, [], ret)
return ret
def dfs(self, nums, index, target, path, ret):
if target == 0 and path not in ret:
ret.append(path)
return #backtracking
elif target < 0 or index >= len(nums):
return #backtracking
# for i in range(index, len(nums)):
# self.dfs(nums, i, target-nums[i], path+[nums[i]], ret)
pick_one = self.dfs(nums, index + 1, target - nums[index], path + [nums[index]], ret)
pick_one_again = self.dfs(nums, index, target - nums[index], path + [nums[index]], ret)
skip_one = self.dfs(nums, index + 1, target, path, ret)
In my opinion, the difference is pruning of tree. Backtracking can stop (finish) searching certain branch in the middle by checking the given conditions (if the condition is not met). However, in DFS, you have to reach to the leaf node of the branch to figure out if the condition is met or not, so you cannot stop searching certain branch until you reach to its leaf nodes.
The difference is: Backtracking is a concept of how an algorithm works, DFS (depth first search) is an actual algorithm that bases on backtracking. DFS essentially is backtracking (it is searching a tree using backtracking) but not every algorithm based on backtracking is DFS.
To add a comparison: Backtracking is a concept like divide and conqueror. QuickSort would be an algorithm based on the concept of divide and conqueror.

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