270. 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.

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.

本题我用了变形题2的思路来做的，时间用的太长了，代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int closestValue(TreeNode root, double target) {
        Stack<TreeNode> succ = new Stack<TreeNode>();
        Stack<TreeNode> pre = new Stack<TreeNode>();
        initalizesucc(succ,root,target);
        initalizepre(pre,root,target);
        if(succ.isEmpty()){
            return pre.peek().val;
        }else if(pre.isEmpty()){
            return succ.peek().val;
        }else{
            double succ_diff = Math.abs((double)succ.peek().val-target);
            double pre_diff = Math.abs((double)pre.peek().val-target);
            if(succ_diff<pre_diff) return succ.peek().val;
            else return pre.peek().val;
        }
    }
    public void initalizesucc(Stack<TreeNode> stack,TreeNode root,double target){
        while(root!=null){
            if(root.val==target){
                stack.push(root);
                break;
            }else if(root.val<target){
                root = root.right;
            }else{
                stack.push(root);
                root = root.left;
            }
        }
    }
    public void initalizepre(Stack<TreeNode> stack,TreeNode root,double target){
        while(root!=null){
            if(root.val==target){
                stack.push(root);
                break;
            }else if(root.val<target){
                stack.push(root);
                root = root.right;
            }else{
                root = root.left;
            }
        }
    }
}

本题还有更为简单的办法，思路是使用递归来做，找出与target左面和右面最近的两个值，返回差值较小者。代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int closestValue(TreeNode root, double target) {
        int a = root.val;
        TreeNode child = target<a?root.left:root.right;
        if(child==null) return a;
        int b = closestValue(child,target);
        return Math.abs(a-target)<Math.abs(b-target)?a:b;
    }
}