Question

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：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。

# 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

/**
 * 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);
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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。

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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;
};