270. 最接近的二叉搜索树值

English Version

题目描述

给你二叉搜索树的根节点 root 和一个目标值 target ，请在该二叉搜索树中找到最接近目标值 target 的数值。如果有多个答案，返回最小的那个。

 

示例 1：

输入：root = [4,2,5,1,3], target = 3.714286
输出：4

示例 2：

输入：root = [1], target = 4.428571
输出：1

 

提示：

• 树中节点的数目在范围 [1, 104] 内
• 0 <= Node.val <= 109
• -109 <= target <= 109

解法

二分查找。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def closestValue(self, root: Optional[TreeNode], target: float) -> int:
        ans, mi = root.val, inf
        while root:
            t = abs(root.val - target)
            if t < mi:
                mi = t
                ans = root.val
            if root.val > target:
                root = root.left
            else:
                root = root.right
        return ans

Java

/**
 * 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) {
        int ans = root.val;
        double mi = Double.MAX_VALUE;
        while (root != null) {
            double t = Math.abs(root.val - target);
            if (t < mi) {
                mi = t;
                ans = root.val;
            }
            if (root.val > target) {
                root = root.left;
            } else {
                root = root.right;
            }
        }
        return ans;
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} target
 * @return {number}
 */
var closestValue = function (root, target) {
    let ans = root.val;
    let mi = Number.MAX_VALUE;
    while (root) {
        const t = Math.abs(root.val - target);
        if (t < mi) {
            mi = t;
            ans = root.val;
        }
        if (root.val > target) {
            root = root.left;
        } else {
            root = root.right;
        }
    }
    return ans;
};

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int closestValue(TreeNode* root, double target) {
        int ans = root->val;
        double mi = INT_MAX;
        while (root) {
            double t = abs(root->val - target);
            if (t < mi) {
                mi = t;
                ans = root->val;
            }
            if (root->val > target)
                root = root->left;
            else
                root = root->right;
        }
        return ans;
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func closestValue(root *TreeNode, target float64) int {
	ans := root.Val
	mi := math.MaxFloat64
	for root != nil {
		t := math.Abs(float64(root.Val) - target)
		if t < mi {
			mi = t
			ans = root.Val
		}
		if float64(root.Val) > target {
			root = root.Left
		} else {
			root = root.Right
		}
	}
	return ans
}

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