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

样例

Given root = {1}, target = 4.428571, return 1.

注意事项

• 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.

解题思路1：

非递归，使用一个变量存储遍历到当前节点时的最接近节点的值。然后依照BST性质遍历。

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: the given BST
     * @param target: the given target
     * @return: the value in the BST that is closest to the target
     */
    public int closestValue(TreeNode root, double target) {
        // write your code here
        TreeNode curNode = root;
        int cloVal = root.val;
        
        while(curNode != null){
            if(Math.abs(target - curNode.val) < Math.abs(target - cloVal))
                cloVal = curNode.val;
            
            if(target > curNode.val){
                curNode = curNode.right;
            }else if(target < curNode.val){
                curNode = curNode.left;
            }
        }
        
        return cloVal;
    }
    
    
}

解题思路2：

递归，思路同上。

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: the given BST
     * @param target: the given target
     * @return: the value in the BST that is closest to the target
     */
    public int closestValue(TreeNode root, double target) {
        // write your code here
        res = root.val;
        
        helper(root, target);
        
        return res;
    }
    
    private int res;
    
    private void helper(TreeNode root, double target){
        if(root == null)
            return;
            
        if(Math.abs(target - root.val) < Math.abs(target - res))
            res = root.val;
            
        if(target > root.val)
            helper(root.right, target);
        else
            helper(root.left, target);
    }
}