270. Closest-Binary-Search-Tree-Value

difficulty: Easy

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

Method One

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int closestValue(TreeNode root, double target) {
        TreeNode cur = root;
​
        int closest = root.val; 
        while(cur != null) {
            closest =  Math.abs(target - cur.val) < Math.abs(target - closest) ?  cur.val : closest;
            if( target < cur.val ) {
                cur = cur.left;
            }else if( target > cur.val) {
                cur = cur.right;
            }else{
                return cur.val;
            }
        }
        return closest;
    }
}
​

2023 写的 应该没看答案至少

class Solution {
    public int closestValue(TreeNode root, double target) {
        double min = Integer.MAX_VALUE + 0.1;
        int minVal = 0;
        TreeNode p = root;
        while (p != null) {
            if (min == Math.abs(p.val - target)) {
                minVal = Math.min(minVal, p.val);
            }
            if (min >  Math.abs(p.val - target)) {
                min = Math.abs(p.val - target);
                minVal = p.val;
            }
            if (p.val < target ) {
                p = p.right;
            } else {
                p = p.left;
            }
        }
        return minVal;
    }
}