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270. 最接近的二叉搜索树值 🔒

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

方法一：中序遍历

我们用一个变量 $mi$ 维护最小的差值，用一个变量 $ans$ 维护答案。初始时 $mi=\infty$, $ans=root.val$。

接下来，进行中序遍历，每次计算当前节点与目标值 $target$ 的差的绝对值 $t$。如果 $t \lt mi$，或者 $t = mi$ 且当前节点的值小于 $ans$，则更新 $mi$ 和 $ans$。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是二叉搜索树的节点数。

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# 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:
        def dfs(root):
            if root is None:
                return
            dfs(root.left)
            nonlocal ans, mi
            t = abs(root.val - target)
            if t < mi:
                mi = t
                ans = root.val
            dfs(root.right)

        ans, mi = root.val, inf
        dfs(root)
        return ans

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/**
 * 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 {
    private int ans;
    private double target;
    private double mi = Double.MAX_VALUE;

    public int closestValue(TreeNode root, double target) {
        this.target = target;
        dfs(root);
        return ans;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        double t = Math.abs(root.val - target);
        if (t < mi) {
            mi = t;
            ans = root.val;
        }
        dfs(root.right);
    }
}

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/**
 * 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;
        function<void(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return;
            }
            dfs(root->left);
            double t = abs(root->val - target);
            if (t < mi) {
                mi = t;
                ans = root->val;
            }
            dfs(root->right);
        };
        dfs(root);
        return ans;
    }
};

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/**
 * 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
    var dfs func(*TreeNode)
    dfs = func(root *TreeNode) {
        if root == nil {
            return
        }
        dfs(root.Left)
        t := math.Abs(float64(root.Val) - target)
        if t < mi {
            mi = t
            ans = root.Val
        }
        dfs(root.Right)
    }
    dfs(root)
    return ans
}

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/**
 * 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 mi = Infinity;
    let ans = root.val;
    const dfs = root => {
        if (!root) {
            return;
        }
        dfs(root.left);
        const t = Math.abs(root.val - target);
        if (t < mi) {
            mi = t;
            ans = root.val;
        }
        dfs(root.right);
    };
    dfs(root);
    return ans;
};

方法二：二分查找

与方法一类似，我们用一个变量 $mi$ 维护最小的差值，用一个变量 $ans$ 维护答案。初始时 $mi=\infty$, $ans=root.val$。

接下来，进行二分查找，每次计算当前节点与目标值 $target$ 的差的绝对值 $t$。如果 $t \lt mi$，或者 $t = mi$ 且当前节点的值小于 $ans$，则更新 $mi$ 和 $ans$。如果当前节点的值大于 $target$，则查找左子树，否则查找右子树。当我们遍历到叶子节点时，就可以结束二分查找了。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 是二叉搜索树的节点数。

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# 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 or (t == mi and root.val < ans):
                mi = t
                ans = root.val
            if root.val > target:
                root = root.left
            else:
                root = root.right
        return ans

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/**
 * 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 || (t == mi && root.val < ans)) {
                mi = t;
                ans = root.val;
            }
            if (root.val > target) {
                root = root.left;
            } else {
                root = root.right;
            }
        }
        return ans;
    }
}

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/**
 * 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 || (t == mi && root->val < ans)) {
                mi = t;
                ans = root->val;
            }
            if (root->val > target) {
                root = root->left;
            } else {
                root = root->right;
            }
        }
        return ans;
    }
};

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/**
 * 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 || (t == mi && root.Val < ans) {
            mi = t
            ans = root.Val
        }
        if float64(root.Val) > target {
            root = root.Left
        } else {
            root = root.Right
        }
    }
    return ans
}

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/**
 * 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 || (t === mi && root.val < ans)) {
            mi = t;
            ans = root.val;
        }
        if (root.val > target) {
            root = root.left;
        } else {
            root = root.right;
        }
    }
    return ans;
};

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