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272. 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.
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.
Follow up:
Assume that the BST is balanced, could you solve it in less than O(n) runtime (where n = total nodes)?
Hint:
- Consider implement these two helper functions:
getPredecessor(N), which returns the next smaller node to N.getSuccessor(N), which returns the next larger node to N.
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用inorder来做 ， bst inorder是有序的。
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ //use inorder it will be a increase arraypublic class Solution { public List<Integer> closestKValues(TreeNode root, double target, int k) { List<Integer> res = new LinkedList<Integer>(); getClosestKValues(res, root, target, k); return res; } public void getClosestKValues(List<Integer> res, TreeNode root, double target, int k){ if(root == null) return; getClosestKValues(res, root.left, target, k); if(res.size() == k){ if(Math.abs(res.get(0) - target) > Math.abs(root.val - target)){ res.remove(0); res.add(root.val); } else return; } else res.add(root.val); getClosestKValues(res, root.right, target, k); }}
272. Closest Binary Search Tree Value II
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