Question

669. 修剪二叉搜索树

English Version

题目描述

给你二叉搜索树的根节点 root ，同时给定最小边界low 和最大边界 high。通过修剪二叉搜索树，使得所有节点的值在[low, high]中。修剪树 不应该 改变保留在树中的元素的相对结构 (即，如果没有被移除，原有的父代子代关系都应当保留)。 可以证明，存在 唯一的答案 。

所以结果应当返回修剪好的二叉搜索树的新的根节点。注意，根节点可能会根据给定的边界发生改变。

 

示例 1：

输入：root = [1,0,2], low = 1, high = 2
输出：[1,null,2]

示例 2：

输入：root = [3,0,4,null,2,null,null,1], low = 1, high = 3
输出：[3,2,null,1]

 

提示：

• 树中节点数在范围 [1, 104] 内
• 0 <= Node.val <= 104
• 树中每个节点的值都是 唯一 的
• 题目数据保证输入是一棵有效的二叉搜索树
• 0 <= low <= high <= 104

解法

方法一：递归

判断 root.val 与 low 和 high 的大小关系：

• 若 root.val 大于 high，说明当前 root 节点与其右子树所有节点的值均大于 high，那么递归修剪 root.left 即可；
• 若 root.val 小于 low，说明当前 root 节点与其左子树所有节点的值均小于 low，那么递归修剪 root.right 即可；
• 若 root.val 在 [low, high] 之间，说明当前 root 应该保留，递归修剪 root.left, root.right，并且返回 root。

递归的终止条件是 root 节点为空。

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个数。

# 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 trimBST(
    self, root: Optional[TreeNode], low: int, high: int
  ) -> Optional[TreeNode]:
    def dfs(root):
      if root is None:
        return root
      if root.val > high:
        return dfs(root.left)
      if root.val < low:
        return dfs(root.right)
      root.left = dfs(root.left)
      root.right = dfs(root.right)
      return root

    return dfs(root)

/**
 * 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 TreeNode trimBST(TreeNode root, int low, int high) {
    if (root == null) {
      return root;
    }
    if (root.val > high) {
      return trimBST(root.left, low, high);
    }
    if (root.val < low) {
      return trimBST(root.right, low, high);
    }
    root.left = trimBST(root.left, low, high);
    root.right = trimBST(root.right, low, high);
    return root;
  }
}

/**
 * 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:
  TreeNode* trimBST(TreeNode* root, int low, int high) {
    if (!root) return root;
    if (root->val > high) return trimBST(root->left, low, high);
    if (root->val < low) return trimBST(root->right, low, high);
    root->left = trimBST(root->left, low, high);
    root->right = trimBST(root->right, low, high);
    return root;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func trimBST(root *TreeNode, low int, high int) *TreeNode {
  if root == nil {
    return root
  }
  if root.Val > high {
    return trimBST(root.Left, low, high)
  }
  if root.Val < low {
    return trimBST(root.Right, low, high)
  }
  root.Left = trimBST(root.Left, low, high)
  root.Right = trimBST(root.Right, low, high)
  return root
}

/**
 * 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 trimBST(root: TreeNode | null, low: number, high: number): TreeNode | null {
  const dfs = (root: TreeNode | null) => {
    if (root == null) {
      return root;
    }
    const { val, left, right } = root;
    if (val < low || val > high) {
      return dfs(left) || dfs(right);
    }
    root.left = dfs(left);
    root.right = dfs(right);
    return root;
  };
  return dfs(root);
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
  pub fn trim_bst(
    mut root: Option<Rc<RefCell<TreeNode>>>,
    low: i32,
    high: i32
  ) -> Option<Rc<RefCell<TreeNode>>> {
    if root.is_none() {
      return root;
    }
    {
      let mut node = root.as_mut().unwrap().borrow_mut();
      if node.val < low {
        return Self::trim_bst(node.right.take(), low, high);
      }
      if node.val > high {
        return Self::trim_bst(node.left.take(), low, high);
      }
      node.left = Self::trim_bst(node.left.take(), low, high);
      node.right = Self::trim_bst(node.right.take(), low, high);
    }
    root
  }
}

/**
 * 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} low
 * @param {number} high
 * @return {TreeNode}
 */
var trimBST = function (root, low, high) {
  function dfs(root) {
    if (!root) {
      return root;
    }
    if (root.val < low) {
      return dfs(root.right);
    }
    if (root.val > high) {
      return dfs(root.left);
    }
    root.left = dfs(root.left);
    root.right = dfs(root.right);
    return root;
  }
  return dfs(root);
};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   struct TreeNode *left;
 *   struct TreeNode *right;
 * };
 */

struct TreeNode* trimBST(struct TreeNode* root, int low, int high) {
  if (!root) {
    return root;
  }
  if (root->val < low) {
    return trimBST(root->right, low, high);
  }
  if (root->val > high) {
    return trimBST(root->left, low, high);
  }
  root->left = trimBST(root->left, low, high);
  root->right = trimBST(root->right, low, high);
  return root;
}

方法二：迭代

我们先循环判断 root，若 root.val 不在 [low, high] 之间，那么直接将 root 置为对应的左孩子或右孩子，循环直至 root 为空或者 root.val 在 [low, high] 之间。

若此时 root 为空，直接返回。否则，说明 root 是一个需要保留的节点。接下来只需要分别迭代修剪 root 的左右子树。

以左子树 node = root.left 为例：

• 若 node.left.val 小于 low，那么 node.left 及其左孩子均不满足条件，我们直接将 node.left 置为 node.left.right；
• 否则，我们将 node 置为 node.left；
• 循环判断，直至 node.left 为空。

右子树的修剪过程与之类似。

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个数。

# 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 trimBST(
    self, root: Optional[TreeNode], low: int, high: int
  ) -> Optional[TreeNode]:
    while root and (root.val < low or root.val > high):
      root = root.left if root.val > high else root.right
    if root is None:
      return None
    node = root
    while node.left:
      if node.left.val < low:
        node.left = node.left.right
      else:
        node = node.left
    node = root
    while node.right:
      if node.right.val > high:
        node.right = node.right.left
      else:
        node = node.right
    return root

/**
 * 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 TreeNode trimBST(TreeNode root, int low, int high) {
    while (root != null && (root.val < low || root.val > high)) {
      root = root.val < low ? root.right : root.left;
    }
    if (root == null) {
      return null;
    }
    TreeNode node = root;
    while (node.left != null) {
      if (node.left.val < low) {
        node.left = node.left.right;
      } else {
        node = node.left;
      }
    }
    node = root;
    while (node.right != null) {
      if (node.right.val > high) {
        node.right = node.right.left;
      } else {
        node = node.right;
      }
    }
    return root;
  }
}

/**
 * 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:
  TreeNode* trimBST(TreeNode* root, int low, int high) {
    while (root && (root->val < low || root->val > high)) {
      root = root->val < low ? root->right : root->left;
    }
    if (!root) {
      return root;
    }
    TreeNode* node = root;
    while (node->left) {
      if (node->left->val < low) {
        node->left = node->left->right;
      } else {
        node = node->left;
      }
    }
    node = root;
    while (node->right) {
      if (node->right->val > high) {
        node->right = node->right->left;
      } else {
        node = node->right;
      }
    }
    return root;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func trimBST(root *TreeNode, low int, high int) *TreeNode {
  for root != nil && (root.Val < low || root.Val > high) {
    if root.Val < low {
      root = root.Right
    } else {
      root = root.Left
    }
  }
  if root == nil {
    return nil
  }
  node := root
  for node.Left != nil {
    if node.Left.Val < low {
      node.Left = node.Left.Right
    } else {
      node = node.Left
    }
  }
  node = root
  for node.Right != nil {
    if node.Right.Val > high {
      node.Right = node.Right.Left
    } else {
      node = node.Right
    }
  }
  return root
}

/**
 * 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} low
 * @param {number} high
 * @return {TreeNode}
 */
var trimBST = function (root, low, high) {
  while (root && (root.val < low || root.val > high)) {
    root = root.val < low ? root.right : root.left;
  }
  if (!root) {
    return root;
  }
  let node = root;
  while (node.left) {
    if (node.left.val < low) {
      node.left = node.left.right;
    } else {
      node = node.left;
    }
  }
  node = root;
  while (node.right) {
    if (node.right.val > high) {
      node.right = node.right.left;
    } else {
      node = node.right;
    }
  }
  return root;
};