#20200120

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

Note:

Given target value is a floating point.
You may assume k is always valid, that is: k ≤ total nodes.
You are guaranteed to have only one unique set of k values in the BST that are closest to the target.
Example:

Input: root = [4,2,5,1,3], target = 3.714286, and k = 2

    4
   / \
  2   5
 / \
1   3

Output: [4,3]

public List<Integer> closestKValues(TreeNode root, double target, int k) {
  List<Integer> res = new ArrayList<>();

  Stack<Integer> s1 = new Stack<>(); // predecessors
  Stack<Integer> s2 = new Stack<>(); // successors

  inorder(root, target, false, s1);
  inorder(root, target, true, s2);
  
  while (k-- > 0) {
    if (s1.isEmpty())
      res.add(s2.pop());
    else if (s2.isEmpty())
      res.add(s1.pop());
    else if (Math.abs(s1.peek() - target) < Math.abs(s2.peek() - target))
      res.add(s1.pop());
    else
      res.add(s2.pop());
  }
  
  return res;
}

// inorder traversal
void inorder(TreeNode root, double target, boolean reverse, Stack<Integer> stack) {
  if (root == null) return;

  inorder(reverse ? root.right : root.left, target, reverse, stack);
  // early terminate, no need to traverse the whole tree
  if ((reverse && root.val <= target) || (!reverse && root.val > target)) return;
  // track the value of current node
  stack.push(root.val); // 1, 2, 3 // 5 4
  inorder(reverse ? root.left : root.right, target, reverse, stack);
}

Design a max stack that supports push, pop, top, peekMax and popMax.

push(x) -- Push element x onto stack.
pop() -- Remove the element on top of the stack and return it.
top() -- Get the element on the top.
peekMax() -- Retrieve the maximum element in the stack.
popMax() -- Retrieve the maximum element in the stack, and remove it. If you find more than one maximum elements, only remove the top-most one

maxStack = [5, 8, 7, 3, 2]
push(5);
push(8);
push(7);
push(3);
push(2);
peekMax() -> 8;
popMax() -> 8;
maxStack = [5, 7, 3, 2]

---------------------------
stack = [5, 8, 7, 3, 2]
maxStack = [5, 8, 8, 8, 8]

class MaxStack {

    Stack<Integer> stack;
    Stack<Integer> maxStack;

    /** initialize your data structure here. */
    public MaxStack() {
        stack = new Stack<>();
        maxStack = new Stack<>();
    }
    
    public void push(int x) {
        int max = maxStack.isEmpty() ? x : maxStack.peek();
        maxStack.push(max > x ? max : x);
        stack.push(x);
    }
    
    public int pop() {
        maxStack.pop();
        return stack.pop();
    }
    
    public int top() {
        return stack.peek();
    }
    
    public int peekMax() {
        return maxStack.peek();
    }
    
    public int popMax() {
        int max = peekMax();
        
        Stack<Integer> bufferStack = new Stack<>();
        while (top() != max) {
            bufferStack.push(pop());
        }
        pop();
        
        while (!bufferStack.isEmpty()) {
            push(bufferStack.pop());
        }
        
        return max;
    }
}