Closest Binary Search Tree Value II

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

Example

Given root = {1}, target = 0.000000, k = 1, return [1].

Challenge

Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)?

Notice

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.

思路

• 跟前一道题很像Closest Binary Search Tree Value II
• 一样的解法也能过，不过这里其实要求更快的方法
• 以前就是遍历一边，再二分法找k相近数值就可以了
• 现在要利用height 平衡，通过去前节点，后节点来实现，比较难
• 九章

最优算法，时间复杂度 O(k + logn)O(k+logn)，空间复杂度 O(logn)O(logn)

实现如下的子函数：

getStack() => 在假装插入 target 的时候，看看一路走过的节点都是哪些，放到 stack 里，用于 iterate
moveUpper(stack) => 根据 stack，挪动到 next node
moveLower(stack) => 根据 stack, 挪动到 prev node
有了这些函数之后，就可以把整个树当作一个数组一样来处理，只不过每次 i++ 的时候要用 moveUpper，i--的时候要用 moveLower

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List<Integer> closestKValues(TreeNode root, double target, int k) {
        List<Integer> values = new ArrayList<>();

        if (k == 0 || root == null) {
            return values;
        }

        Stack<TreeNode> lowerStack = getStack(root, target);
        Stack<TreeNode> upperStack = new Stack<>();
        upperStack.addAll(lowerStack);
        if (target < lowerStack.peek().val) {
            moveLower(lowerStack);
        } else {
            moveUpper(upperStack);
        }

        for (int i = 0; i < k; i++) {
            if (lowerStack.isEmpty() ||
                   !upperStack.isEmpty() && target - lowerStack.peek().val > upperStack.peek().val - target) {
                values.add(upperStack.peek().val);
                moveUpper(upperStack);
            } else {
                values.add(lowerStack.peek().val);
                moveLower(lowerStack);
            }
        }

        return values;
    }

    private Stack<TreeNode> getStack(TreeNode root, double target) {
        Stack<TreeNode> stack = new Stack<>();

        while (root != null) {
            stack.push(root);

            if (target < root.val) {
                root = root.left;
            } else {
                root = root.right;
            }
        }

        return stack;
    }

    public void moveUpper(Stack<TreeNode> stack) {
        TreeNode node = stack.peek();
        if (node.right == null) {
            node = stack.pop();
            while (!stack.isEmpty() && stack.peek().right == node) {
                node = stack.pop();
            }
            return;
        }

        node = node.right;
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
    }

    public void moveLower(Stack<TreeNode> stack) {
        TreeNode node = stack.peek();
        if (node.left == null) {
            node = stack.pop();
            while (!stack.isEmpty() && stack.peek().left == node) {
                node = stack.pop();
            }
            return;
        }

        node = node.left;
        while (node != null) {
            stack.push(node);
            node = node.right;
        }
    }
}