LeetCode 270. Closest Binary Search Tree Value（二叉搜索树最接近值查找）

2019年3月16日 432次阅读 来源: 二叉查找树

原题网址：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.

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.

方法：递归查找。如果当前节点值小于目标值，则结果只可能是当前节点值或者右子树中的值；如果当前节点值大于目标值，则结果只可能是当前节点值或者左子树中的值。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int value;
    private void find(TreeNode root, double target) {
        if (Math.abs(root.val-target) < Math.abs(value-target)) value = root.val;
        if (root.val < target && root.right != null) find(root.right, target);
        if (root.val > target && root.left != null) find(root.left, target);
    }
    public int closestValue(TreeNode root, double target) {
        value = root.val;
        find(root, target);
        return value;
    }
}

优化：

/**
 * 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) {
        if (root.val == target) return root.val;
        if (root.val < target) {
            if (root.right == null) return root.val;
            int right = closestValue(root.right, target);
            if (Math.abs(root.val-target) <= Math.abs(right-target)) return root.val;
            return right;
        } else {
            if (root.left == null) return root.val;
            int left = closestValue(root.left, target);
            if (Math.abs(root.val-target) <= Math.abs(left-target)) return root.val;
            return left;
        }
    }
}

    原文作者：二叉查找树
    原文地址: https://blog.csdn.net/jmspan/article/details/51139874
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
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