Leetcode/各大家 -- 270. Closest Binary Search Tree Value

270. Closest Binary Search Tree Value

https://leetcode.com/problems/closest-binary-search-tree-value/
Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.

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参考： https://segmentfault.com/a/1190000003797291 

1. recursion

Complexity: 时间 O(logN) 空间 O(H)

思路

根据二叉树的性质，我们知道当遍历到某个根节点时，最近的那个节点要么是在子树里面，要么就是根节点本身。所以我们根据这个递归，返回子树中最近的节点，和根节点中更近的那个就行了。

public class Solution {
    public int closestValue(TreeNode root, double target) {
        // 选出子树的根节点
        TreeNode kid = target < root.val ? root.left : root.right;
        // 如果没有子树，也就是递归到底时，直接返回当前节点值
        if(kid == null) return root.val;
        // 找出子树中最近的那个节点
        int closest = closestValue(kid, target);
        // 返回根节点和子树最近节点中，更近的那个节点
        return Math.abs(root.val - target) < Math.abs(closest - target) ? root.val : closest;
    }
}

2.Iterative
We the node is in tree so traveling the BST and keep a value closet to target

Complexity: O(lgN)

public class Solution {
    public int closestValue(TreeNode root, double target) {
        int closest = root.val;
        while(root!=null){
            //if current root.val is closer, update closest 
            closest = Math.abs(closest-target)<Math.abs(root.val-target)?closest:root.val;
            //do binary search
            root = target<root.val?root.left:root.right;
        }
        return closest;
    }
}