Skip to content

Latest commit

 

History

History
490 lines (422 loc) · 12 KB

README.md

File metadata and controls

490 lines (422 loc) · 12 KB
Raw
comments difficulty edit_url tags
true
简单
树
深度优先搜索
二叉搜索树
二分查找
二叉树

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

方法一：递归

我们定义一个递归函数 $\text{dfs}(node)$，表示从当前节点 $node$ 开始，寻找最接近目标值 $target$ 的节点。我们可以通过比较当前节点的值与目标值的差的绝对值，来更新答案，如果目标值小于当前节点的值，我们就递归地搜索左子树，否则我们递归地搜索右子树。

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。

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:
        def dfs(node: Optional[TreeNode]):
            if node is None:
                return
            nxt = abs(target - node.val)
            nonlocal ans, diff
            if nxt < diff or (nxt == diff and node.val < ans):
                diff = nxt
                ans = node.val
            node = node.left if target < node.val else node.right
            dfs(node)

        ans = 0
        diff = inf
        dfs(root)
        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 {
    private int ans;
    private double target;
    private double diff = Double.MAX_VALUE;

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

    private void dfs(TreeNode node) {
        if (node == null) {
            return;
        }
        double nxt = Math.abs(node.val - target);
        if (nxt < diff || (nxt == diff && node.val < ans)) {
            diff = nxt;
            ans = node.val;
        }
        node = target < node.val ? node.left : node.right;
        dfs(node);
    }
}

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 diff = INT_MAX;
        function<void(TreeNode*)> dfs = [&](TreeNode* node) {
            if (!node) {
                return;
            }
            double nxt = abs(node->val - target);
            if (nxt < diff || (nxt == diff && node->val < ans)) {
                diff = nxt;
                ans = node->val;
            }
            node = target < node->val ? node->left : node->right;
            dfs(node);
        };
        dfs(root);
        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
	diff := math.MaxFloat64
	var dfs func(*TreeNode)
	dfs = func(node *TreeNode) {
		if node == nil {
			return
		}
		nxt := math.Abs(float64(node.Val) - target)
		if nxt < diff || (nxt == diff && node.Val < ans) {
			diff = nxt
			ans = node.Val
		}
		if target < float64(node.Val) {
			dfs(node.Left)
		} else {
			dfs(node.Right)
		}
	}
	dfs(root)
	return ans
}

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function closestValue(root: TreeNode | null, target: number): number {
    let ans = 0;
    let diff = Number.POSITIVE_INFINITY;

    const dfs = (node: TreeNode | null): void => {
        if (!node) {
            return;
        }

        const nxt = Math.abs(target - node.val);
        if (nxt < diff || (nxt === diff && node.val < ans)) {
            diff = nxt;
            ans = node.val;
        }

        node = target < node.val ? node.left : node.right;
        dfs(node);
    };

    dfs(root);
    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 = 0;
    let diff = Infinity;

    const dfs = node => {
        if (!node) {
            return;
        }

        const nxt = Math.abs(target - node.val);
        if (nxt < diff || (nxt === diff && node.val < ans)) {
            diff = nxt;
            ans = node.val;
        }

        node = target < node.val ? node.left : node.right;
        dfs(node);
    };

    dfs(root);
    return ans;
};

方法二：迭代

我们可以将递归函数改写为迭代的形式，使用一个循环来模拟递归的过程。我们从根节点开始，判断当前节点的值与目标值的差的绝对值是否小于当前的最小差，如果是，我们就更新答案。然后根据目标值与当前节点的值的大小关系，决定向左子树还是右子树移动。当我们遍历到空节点时，循环结束。

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

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, diff = root.val, inf
        while root:
            nxt = abs(root.val - target)
            if nxt < diff or (nxt == diff and root.val < ans):
                diff = nxt
                ans = root.val
            root = root.left if target < root.val else 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 diff = Double.MAX_VALUE;
        while (root != null) {
            double nxt = Math.abs(root.val - target);
            if (nxt < diff || (nxt == diff && root.val < ans)) {
                diff = nxt;
                ans = root.val;
            }
            root = target < root.val ? root.left : 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 diff = INT_MAX;
        while (root) {
            double nxt = abs(root->val - target);
            if (nxt < diff || (nxt == diff && root->val < ans)) {
                diff = nxt;
                ans = root->val;
            }
            root = target < root->val ? root->left : 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
	diff := math.MaxFloat64
	for root != nil {
		nxt := math.Abs(float64(root.Val) - target)
		if nxt < diff || (nxt == diff && root.Val < ans) {
			diff = nxt
			ans = root.Val
		}
		if float64(root.Val) > target {
			root = root.Left
		} else {
			root = root.Right
		}
	}
	return ans
}

TypeScript

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function closestValue(root: TreeNode | null, target: number): number {
    let ans = 0;
    let diff = Number.POSITIVE_INFINITY;

    while (root) {
        const nxt = Math.abs(root.val - target);
        if (nxt < diff || (nxt === diff && root.val < ans)) {
            diff = nxt;
            ans = root.val;
        }
        root = target < root.val ? root.left : 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 diff = Infinity;
    while (root) {
        const nxt = Math.abs(root.val - target);
        if (nxt < diff || (nxt === diff && root.val < ans)) {
            diff = nxt;
            ans = root.val;
        }
        root = target < root.val ? root.left : root.right;
    }
    return ans;
};