#binary trees

Recommended Book (1) Introduction to Algorithms --Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Clifford Stein .

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# Given a binary tree where all nodes are either 0 or 1, prune the tree so that subtrees containing all 0s are removed. # For example, given the following tree: # 0 # / \ # 1 0 # / \ # 1 0 # / \ # 0 0 # should be pruned to: # 0 # / \ # 1 0 # / # 1 # We do not remove the tree at the root or its left child because it still has a 1 as a descendant. # 0 # / \ # 0 1 # \ \ # 1 0 # / \ # 0 0 # \ # 1 # pruned: # 0 # / \ # 0 1 # \ # 1 # \ # 0 # \ # 1 # 0 # left not none, right not none # / \ # > 0 1 < # left none, is a one # \ \ # > 1 0 < # is a 1, 0.right is none and 0.left is none -> can be pruned # / \ # > 0 0 < # 0.right is none and 0.left is none -> can be pruned, 0.right not none # \ # 1 < is a 1 # we may be able to do it via a BFS/level order search and on each level check each current node # if current node.left and current node.right is none and current node.val == 0 then remove this node # For each node, first the node is visited and then it’s child nodes are put in a FIFO queue. #printLevelorder(tree) # 1) Create an empty queue q # 2) temp_node = root /*start from root*/ # 3) Loop while temp_node is not NULL # a) print temp_node->data. # b) Enqueue temp_node’s children (first left then right children) to q # c) Dequeue a node from q and assign it’s value to temp_node # q = 0 # 0 <- cant be pruned # check temp for left.none & right.none # enqueue left & right # q = 0,1 # 0.left = none but 0 .right = 1 <- cant be pruned # q = 1,1 # 1 is 1 cant be pruned # q = 1,0 # 1 is 1 cant be pruned # q = 0,0,0 # 0.left and 0.right are none <- remove this node # q = 0,0 # 0.left and 0.right are none <- remove this node # q = 0 # 0.right = 1 <-cant be pruned # q = 1 # 1 is 1 # reached end import queue as queue def prune(root): q = queue() q.append(root) while len(queue) > 0: tempNode = q.dequeue() if tempNode.left is not None: q.append(tempNode.left) if tempNode.right is not None: q.append(tempNode.right) if tempNode.val == 0 and tempNode.right is None and tempNode.left is None: pass # however we need to reconstruct the tree after pruning which would be tedious, it wouldn't affect runtime complexity but it may be seen as inefficent # A better way of doing this could be to Prune children of the tree recursively. The only decisions at each node are whether to prune the left child or the right child. # Algorithm # We'll use a function containsOne(node) that does two things: it tells us whether the subtree at this node contains a 1, and it also prunes all subtrees not containing 1. # If for example, node.left does not contain a one, then we should prune it via node.left = null. # Also, the parent needs to be checked. If for example the tree is a single node 0, the answer is an empty tree. def pruneRecursively(root): if root.val == 0 and root.left is None and root.right is None: return None root.left = pruneRecursively(root.left) root.right = pruneRecursively(root.right) return root

As always, it’s a day before the exam and I wish I can kick my nervousness to Andromeda, but unfortunately my kicking poweress is not that much— But that’s fine, we’ll all just kick the test’s ass instead. Good luck everyone!

Coding coding coding. But finally finished my “project’s” code... So many things to fix though :/ I need to figure out how to free my array without causing a segmentation fault. But at least it works?

(This is slow above catalan(10) though, so here's a memoised version. Then if you want to run above catalan(250) you'll need to adjust options(expressions=5e3) to something higher; otherwise the nesting of calls goes too deep.)