Leetcode solution 270: Closest Binary Search Tree Value

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发布于 2019-05-13 16:12:58

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发布于 2019-05-13 16:12:58

文章被收录于专栏：包子铺里聊IT

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Leetcode solution 270: Closest Binary Search Tree Value

https://blog.baozitraining.org/2019/04/leetcode-solution-270-closest-binary.html

https://youtu.be/GcrvRcEM9Hs

Problem Statement

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

Note:

Given target value is a floating point.
You are guaranteed to have only one unique value in the BST that is closest to the target.

Example:

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

    4
   / \
  2   5
 / \
1   3

Output: 4

Problem link

Video Tutorial

You can find the detailed video tutorial here

Thought Process

This is a very easy problem as long as you are familiar with Binary Search Tree. In order to find the value that is closest to the given target, we can traverse the tree and keep a record of the minimum delta. Of course we can traverse the entire tree but we can definite do it better by always branch out either left or right given the target and root node value. For example, if current node is 10, and target is 9, the next possible smaller value must exist only on the left part of or with the current node

     10
   /   \
  7     11

Solutions

Recursion solution

	private double curMinDelta = Double.MAX_VALUE;
	private int smallestValue = 0;

	public int closestValue(TreeNode root, double target) {
	this.searchMin(root, target);
	return this.smallestValue;
	}

	private void searchMin(TreeNode root, double target) {
	if (root == null) {
	return;
	}

	if (target == root.val) {
	this.smallestValue = root.val;
	return;
	}

	double delta = Math.abs(target - root.val);
	if (delta < this.curMinDelta) {
	smallestValue = root.val;
	this.curMinDelta = delta;
	}

	if (target < root.val) {
	this.searchMin(root.left, target);
	} else {
	this.searchMin(root.right, target);
	}
	}

view rawCloestBinarySearchTreeValue.java hosted with ❤ by GitHub

Time Complexity: assuming tree size is N,O(lgN) since we are doing binary search Space Complexity:O(1) if not considering recursion function stacks.

Iterative solution

public int closestValue(TreeNode root, double target) {
    int ret = root.val;   
    while(root != null){
        if(Math.abs(target - root.val) < Math.abs(target - ret)){
            ret = root.val;
        }      
        root = root.val > target? root.left: root.right;
    }     
    return ret;
}