Let m, n be two distinct nodes in a tree T. Show that exactly one of the following four cases holds.
(a) m is to the left of n.
(b) m is a proper ancestor of n.
(c) m is to the right of n.
(d) m is a proper descendent of n.

I am not able to start it. It seems intuitive as per the structure of tree. There are many questions like this in my assignment. Any hints for these type of questions would be helpful.

How to prove these type of questions for Data Structures?

i want to code all binary tree orders in single code (preorder,postorder,inorder) in O(N) time complexity and O(N) space , using single stack and without recursion. 

can we code inorder,preorder and postorder in single code in python ? without using recursion

It has a conceptual question about tree data structure. Can we have some child data , then define some parents and assign them to selected child ones and build a tree data structure? 

Define child before parent in a tree data structure

I was exploring about AVL trees and found this answer where the person says that the look ups in AVL is faster as compared to insertion or deletion.

Though for all those 3 processes the worst time complexity is still Logarithmic.

Is it because of the rotation or balancing involved during insertion ?

Avl tree look up vs insertion

What is the difference between complete binary tree, almost complete binary tree and perfect binary tree?

In my institute professor told they are different but on certain sites they have been merged almost complete binary tree and (complete binary tree and perfect binary tree as same). Now I am very much confused so had to ask world's largest tech community.

What is the difference between complete binary tree, almost complete binary tree and perfect binary tree? 

What is the difference between a member variable and a local variable?

I created a generic Binary Node Class, as well as a generic Binary Tree class that uses said Node Class. Now I would like to use these classes to store objects with multiple values that are sorted by one of those attributes.

I tried around a lot and I'm lost. I can't seem to find a solution anywhere. Where to store data that is not the key value?

Example object: "Car" with values "brand name" and "construction year", cars are to be sorted by construction year

How to store Objects with a key value in Binary Trees?

For the purpose of memory performance optimizations while building a Huffman tree I would like to pre-allocate the necessary memory for its nodes.

Is there a way to calculate the maximum number of nodes (internal nodes plus leafs)?

Input for the calculation should be the table of symbols and their probabilities/frequencies. I would like to avoid a simulated tree building run. Instead it should be a plain calculation that must not give the actual/optimal number of nodes but a reliable maximum.

Maximum number of nodes in a Huffman tree

I want to code all binary tree orders in single code (preorder,postorder,inorder) in O(N) time complexity and O(N) space , using single stack and without recursion.

I was solving some problems on Binary tree and I got stuck in this question https://www.geeksforgeeks.org/sum-nodes-longest-path-root-leaf-node/ I am using python to solve the question I understood the logic of the solution given on the link but my question is how did the value of maxSum change in the SumOfLongRootToLeafPathUtil(root) function when nothing is returned from SumOfLongRootToLeafPath() function how is the original value of the varaible change please help ps:Please refer to the python code given in the link 

How is the value returned from the function SumOfLongRootToLeafPath

I'm trying to figure out a function in c++ to calculate the sums of all the levels of a binary tree and then checking if those sums are all equal, and returning TRUE if they are.

So for example, if the first node is 10, the sum of its children has to be 10, and the sum of their children has to be 10, and so on...

I'm having truble comparing the sums without the recursions getting in the way. And the comparison has to be done in the function itself, not the main.

if you have defined the Node data as follows

     TreeNode() : val(0), left(nullptr), right(nullptr) {}

     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}

     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}

Then you can perform level order traversal without recursion as follows.

3. now traverse children until the queue is empty

4. while traversing, if you find left or right children you have to push them again into the queue

Here's the basic code that does what you wanted

        int sum = root->val; //save the current root value

            int s = 0; //keep track of the next levels sum

            len = q.size(); //size of queue is number of elements at that level

            // break if sum is not equal to sum of node values in current level

I think I give u a satisfactory answer to your question.

Checking if the sums of the levels of a Binary Tree are equal

I generated a decision tree which is in if else form. Is there any way to convert it into a tree form in sklearn so that i could use various other functions like accuracyscore() on the tree? 

Is there any way to convert rules generated to a tree in sklearn?

I have got the subsequent code to assemble a binary tree from a listing. But I do not understand the order of the output. If i enter [1,2,3] the output is [2,1,3]. I notion it has to be [1,3,2] can all people help me explain it... I can not think it via.

//    def __init__(self, val=0, left=None, right=None):

            root.left=constructBST(l,mid-1)

            root.right=constructBST(mid+1,r)

        return constructBST(0,len(nums)-1)

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