The problem here is you return a reference of a local variable to outer scope but the local variable will be deleted if it goes out of inner scope. P/s: great content :)
@@sahilchoudhary1473 In the video, Errichto got a runtime error while he is trying to return the address of the root node (a local variable) to the outer scope. The reason is that the root node which is a local variable went out of scope (the function is returned) and it's automatically released (destroyed) so you no longer access the returned value from the caller function.
@@bacuongcao2170 no I don't think this is the case because no matter if its local or global variable objects go to the heap, its stack that gets destroyed after method exit but variables stored in heap remain in the heap. so the method is just returning the reference to the object in the heap, which still exists. correct if I am wrong I am telling this from java point of view.
Hi Errichto . On Behalf of the coding community I would like to request if you could continue the LeetCode Challenge so that we can learn from it. It will be helpful for all of us as we had our previous Leetcode Monthly Challenge. Thanks
Thank you very much, didn't understand the code of O(n) solution, but drawing at the end of the video helped me to get the idea, now I can try to code it myself
for my solution i used a stack to save the visited nodes, first i start by adding the root to the stack and for each element from the sequence if it's less than the top i do top->left = current. otherwise, i pop from the while the element values are small than the current value, eventually, there will be at least an element in the stack (at the end i add current to the right of the last element that had a value smaller than current)
The linear solution is very elegant. For some reason, my brain worked better with a stack than a recursion. I put on stack the first root (the result) and then if the next value is less than the stack top I put it on the stack, if not I pop from the stack to the moment it's either empty or the top is not less than the value and put again the new node on the stack.
How can I be like you ,what course should I do ,plz sir help me plz suggest me something ! That can I follow in daily basis to improve coding 🙏🙏🙏🙏🙏♥️♥️♥️♥️
I implemented it using iterative dfs, with upper and lower limit for each node. But it’s slower then your approach haha Ps just realised lower limit is not useful here
As we were only asked to return BST root which will be formed from preorder traversal we can just write a insert() function and it works in O(n) in worst case.
"new" keyword makes the value stored in heap memory. So even if the variable holding it gets out of scope it won't get destroyed until manually deleted or the whole program is terminated.
Write c program to implement binary search tree by inserting in following 6,2,8,10,4,3,7 write c routin for find operation min.max.delete please solve the. Problem
can such a method also be used for recreating BST from postorder, inorder traversal? what extra thing is needed for recreating tree from level-order traversal?
You're an amazing competitive programmer, but probably you should learn the c++ language a bit. For example, it was surprising to me that you didn't know about local variable's lilfetime and that it is an issue to return reference to it as it is already destroyed.
I have written a python3 implementation...O(N), it basically remembers the minimum possible value and maximum possible value at a given node and decides whether the current node fits in the position. check it out here: pastebin.com/MrNg1EiF
