Leetcode-closest Binary Search Tree Value II

Given a non-empty binary search tree and a target value, find K values in the BST that is closest to the target.

Note:

• Given target value is a floating point.
• Assume k is always valid, which is: k ≤total nodes.
• You is guaranteed to a unique set of K values in the BST is closest to the target.

Follow up:
Assume that the BST are balanced, could you solve it in less than O(n) runtime (where n = Total nodes)?

Analysis:

Use inorder Traverse, put all predecessors to a stack, for every successor, put all Pres that have smaller gap than that Successor into Reslist and then put this successor into Reslist.

Solution:

/*** Definition for a binary tree node. * public class TreeNode {* int val; * TreeNode left; * TreeNode rig Ht * TreeNode (int x) {val = x;} }*/ Public classSolution { PublicList<integer> closestkvalues (TreeNode root,DoubleTargetintk) {Stack<Integer> pres =NewStack<integer>(); LinkedList<Integer> reslist =NewLinkedlist<integer>();        Closestkvaluesrecur (root,target,k,pres,reslist); //If Not enough in reslist, put to pres into reslist. This is because successor is too little.         while(Reslist.size () <k &&!Pres.empty ())        {Reslist.addfirst (Pres.pop ()); }        returnreslist; }     Public voidClosestkvaluesrecur (TreeNode Curnode,DoubleTargetintK, stack<integer> pres, linkedlist<integer>reslist) {        if(Curnode = =NULL)return; if(Reslist.size () ==k)return; //inorder Traverse.closestkvaluesrecur (curnode.left,target,k,pres,reslist); //Check Curnode        if(Curnode.val >=target) {             while(Reslist.size () <k &&!pres.empty () && Target-pres.peek () < curnode.val-target)            {Reslist.addfirst (Pres.pop ()); }            if(Reslist.size () <k)            {reslist.addlast (curnode.val); } Else {                return; }        } Else{Pres.push (curnode.val);    } closestkvaluesrecur (Curnode.right,target,k,pres,reslist); }}

Leetcode-closest Binary Search Tree Value II