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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 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)?
Analysis:
Use inorder traverse, put all predecessors into a stack, for every successor, put all pres that has 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 right; * TreeNode(int x) { val = x; } * } */public class Solution { public List<Integer> closestKValues(TreeNode root, double target, int k) { Stack<Integer> pres = new Stack<Integer>(); LinkedList<Integer> resList = new LinkedList<Integer>(); closestKValuesRecur(root,target,k,pres,resList); // If not enough in resList, put more pres into resList. This is because successor is too little. while (resList.size()<k && !pres.empty()){ resList.addFirst(pres.pop()); } return resList; } public void closestKValuesRecur(TreeNode curNode, double target, int k, 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
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