Find K nodes in BST that are closest to a value key (Part I). | Algorithms…..A lot of fun…..
There is a BST while each node has distinct values, then for a given value key, find out K nodes in this BST such that their values are closest to key.
Solution:
It seems to be a very easy question, all we need to do is to do a pre-order traversal and store all the nodes in this BST to an array, and then find out the K numbers that are the nearest to key. However, This approach takes O(n) time.
It is introduced in this blog an O(log(n)) approach with the assumption that the node in BST has “Parent” pointer.
https://csjobinterview.wordpress.com/2012/06/29/find-k-nodes-in-bst-that-are-closest-to-a-value-key-part-ii/
This blog provide an O(log(n)) approach without the help of the “Parent” pointers. It uses extra addition space to store a stack, which stores all the parent nodes when travelling at a given node.
Read full article from Find K nodes in BST that are closest to a value key (Part I). | Algorithms…..A lot of fun…..
There is a BST while each node has distinct values, then for a given value key, find out K nodes in this BST such that their values are closest to key.
Solution:
It seems to be a very easy question, all we need to do is to do a pre-order traversal and store all the nodes in this BST to an array, and then find out the K numbers that are the nearest to key. However, This approach takes O(n) time.
It is introduced in this blog an O(log(n)) approach with the assumption that the node in BST has “Parent” pointer.
1) Start from the root node to find out the node (denoted as the closestNode) whose value is the closest to key.
2) Generate two functions the nextLargerNodeInBST() and nextSmallerNodeInBST(), using which to find the next larger node of the closestNode, denoted as nextLarger, and find its next smaller node, denoted as nextSmaller.
3) Compare nextSmaller to nextLarger by their difference to key, and update them appropriately until K nodes are selected.
2) Generate two functions the nextLargerNodeInBST() and nextSmallerNodeInBST(), using which to find the next larger node of the closestNode, denoted as nextLarger, and find its next smaller node, denoted as nextSmaller.
3) Compare nextSmaller to nextLarger by their difference to key, and update them appropriately until K nodes are selected.
Step 1) costs O(log(n)).
Step 3) It takes at most O(log(n)) to find out next point, and each node in BST is checked at most twice in the entire searching, therefore, the overall time complexity is max {O(K), O(log(n))} = O(log(n))
Step 3) It takes at most O(log(n)) to find out next point, and each node in BST is checked at most twice in the entire searching, therefore, the overall time complexity is max {O(K), O(log(n))} = O(log(n))
public static List<Tree> findKClosestNodeInBST(Tree root, int key, int k){ List<Tree> list = new List<Tree>(); Tree closestNode = findTreeWithKeyNearestToTheKey(root, key); k--; list.Add(closestNode); Tree nextlarger = nextLargerNodeInBST(closestNode); Tree nextSmaller = nextSmallerNodeInBST(closestNode); while (k > 0) { if (nextlarger == null && nextSmaller == null) throw new StackOverflowException(); else if (nextlarger != null && nextSmaller != null) { if (Math.Abs(nextlarger.node - key) >= Math.Abs(nextSmaller.node - key)) { list.Add(nextSmaller); k--; nextSmaller = nextSmallerNodeInBST(nextSmaller); } else { list.Add(nextlarger); k--; nextlarger = nextLargerNodeInBST(nextlarger); } } else if (nextlarger != null) { list.Add(nextlarger); k--; nextlarger = nextLargerNodeInBST(nextlarger); } else { list.Add(nextSmaller); k--; nextSmaller = nextSmallerNodeInBST(nextSmaller); } } return list;}//find out the node that is closed to the key, step 1)public static Tree findTreeWithKeyNearestToTheKey(Tree root, int key){ Tree desiredRoot = root; int diff = Math.Abs(root.node - key); while (root != null) { if (diff > Math.Abs(root.node - key)) { diff = Math.Abs(root.node - key); desiredRoot = root; } if (root.node > key) root = root.leftChild; else if (root.node < key) root = root.rightChild; else return root; } return desiredRoot;}//step 2) find its next larger node in BSTpublic static Tree nextLargerNodeInBST(Tree current){ if (current.rightChild != null) { Tree nextTree = current.rightChild; while (nextTree.leftChild != null) nextTree = nextTree.leftChild; return nextTree; } else { while (current.parentTree!=null) { if (current != current.parentTree.rightChild) return current.parentTree; else { while(current.parentTree!=null&¤t==current.parentTree.rightChild) current = current.parentTree; return current.parentTree; } } return null; }}//step 2) find its next smaller node in BSTpublic static Tree nextSmallerNodeInBST(Tree current){ if (current.leftChild != null) { Tree nextTree = current.leftChild; while (nextTree.rightChild != null) nextTree = nextTree.rightChild; return nextTree; } else { while (current.parentTree != null) { if (current == current.parentTree.rightChild) return current.parentTree; else { while (current.parentTree != null && current == current.parentTree.leftChild) current = current.parentTree; return current.parentTree; } } return null; }}This blog provide an O(log(n)) approach without the help of the “Parent” pointers. It uses extra addition space to store a stack, which stores all the parent nodes when travelling at a given node.
1) Start from the root node to find out the node (denoted as the closestNode) whose value is closest to key, store all the passed middle-level nodes in a stack where stack.peek() is the closeNode. In this case, the return value should be a stack instead of a single node.
2) Generate two functions the nextLargerNodeInBSTIterative() and nextSmallerNodeInBSTIterative(), by the help of the stack and based on closestNode find its next larger node, denoted as nextLarger, and find its next smaller node, denoted as nextSmaller.
3) Compare nextSmaller to nextLarger by their difference to key, and update them appropriately until K nodes are selected.
2) Generate two functions the nextLargerNodeInBSTIterative() and nextSmallerNodeInBSTIterative(), by the help of the stack and based on closestNode find its next larger node, denoted as nextLarger, and find its next smaller node, denoted as nextSmaller.
3) Compare nextSmaller to nextLarger by their difference to key, and update them appropriately until K nodes are selected.
Step 1) costs O(log(n)), with space complexity O(log(n)).
Step 3) In the worse case, it takes O(log(n)) to find its next larger node or its next smaller node, however, each node in BST is check at most once. Therefore the time complexity should be max{O(log(n)), O(K)} = O(log(n)).
Step 3) In the worse case, it takes O(log(n)) to find its next larger node or its next smaller node, however, each node in BST is check at most once. Therefore the time complexity should be max{O(log(n)), O(K)} = O(log(n)).
public static List findKClosestNodeInBSTIterative(Tree root, int key, int k){ List list = new List(); Stack stack = findTreeWithKeyNearestToTheKeyIterative(root, key); Tree closestNode = stack.Pop(); k--; list.Add(closestNode); Stack stackForSmaller = new Stack(), stackForGreater = new Stack(), temp = new Stack(); //copy the current stack to two stacks //so they can be used in findnext functions. while (stack.Count > 0) temp.Push(stack.Pop()); while (temp.Count > 0) { Tree tempTree = temp.Pop(); stackForSmaller.Push(tempTree); stackForGreater.Push(tempTree); } Tree nextlarger = nextLargerNodeInBSTIterative(closestNode,stackForGreater); Tree nextSmaller = nextSmallerNodeInBSTIterative(closestNode,stackForSmaller); while (k > 0) { if (nextlarger == null && nextSmaller == null) throw new StackOverflowException(); else if (nextlarger != null && nextSmaller != null) { if (Math.Abs(nextlarger.node - key) >= Math.Abs(nextSmaller.node - key)) { list.Add(nextSmaller); k--; nextSmaller = nextSmallerNodeInBSTIterative(nextSmaller, stackForSmaller); } else { list.Add(nextlarger); k--; nextlarger = nextLargerNodeInBSTIterative(nextlarger, stackForGreater); } } else if (nextlarger != null) { list.Add(nextlarger); k--; nextlarger = nextLargerNodeInBSTIterative(nextlarger, stackForGreater); } else { list.Add(nextSmaller); k--; nextSmaller = nextSmallerNodeInBSTIterative(nextSmaller, stackForSmaller); } } return list;} public static Stack findTreeWithKeyNearestToTheKeyIterative(Tree root, int key){ Stack stack = new Stack(); Tree desiredRoot = root; int diff = Math.Abs(root.node - key); while (root != null) { stack.Push(root); if (diff > Math.Abs(root.node - key)) { diff = Math.Abs(root.node - key); desiredRoot = root; } if (root.node > key) root = root.leftChild; else if (root.node < key) root = root.rightChild; else return stack; } while (stack.Peek() != desiredRoot) stack.Pop(); return stack;}//step 2) find its next larger node public static Tree nextLargerNodeInBSTIterative(Tree current, Stackstack){ if (current.rightChild != null) { Tree nextTree = current.rightChild; while (nextTree.leftChild != null) { stack.Push(nextTree); nextTree = nextTree.leftChild; } return nextTree; } else { if (stack.Count > 0) { Tree tempTree = stack.Pop(); while (tempTree != null) { if (tempTree.node > current.node) break; else tempTree = stack.Count > 0 ? stack.Pop() : null; } return tempTree; } else return null; }}//step 2) find its next smaller nodepublic static Tree nextSmallerNodeInBSTIterative(Tree current, Stack stack){ if (current.leftChild != null) { Tree nextTree = current.leftChild; while (nextTree.rightChild != null) { stack.Push(nextTree); nextTree = nextTree.rightChild; } return nextTree; } else { if (stack.Count > 0) { Tree tempTree = stack.Pop(); while (tempTree != null) { if (tempTree.node < current.node) break; else tempTree = stack.Count>0?stack.Pop():null; } return tempTree; } else return null; }}Read full article from Find K nodes in BST that are closest to a value key (Part I). | Algorithms…..A lot of fun…..
