Skip to content

Latest commit

 

History

History
82 lines (71 loc) · 2.16 KB

669. Trim a Binary Search Tree.md

File metadata and controls

82 lines (71 loc) · 2.16 KB
Raw

669. Trim a Binary Search Tree

leetcode Daily Challenge on Feburary 2nd, 2021.

Difficulty : Medium

Related Topics : Tree、Recursion

Given the root of a binary search tree and the lowest and highest boundaries as low and high, trim the tree so that all its elements lies in [low, high]. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a unique answer.

Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.

Example 1:

1

Input: root = [1,0,2], low = 1, high = 2
Output: [1,null,2]

Example 2:

2

Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
Output: [3,2,null,1]

Example 3:

Input: root = [1], low = 1, high = 2
Output: [1]

Example 4:

Input: root = [1,null,2], low = 1, high = 3
Output: [1,null,2]

Example 5:

Input: root = [1,null,2], low = 2, high = 4
Output: [2]

Constraints:

  • The number of nodes in the tree in the range [1, 10^4].
  • 0 <= Node.val <= 10^4
  • The value of each node in the tree is unique.
  • root is guaranteed to be a valid binary search tree.
  • 0 <= low <= high <= 10^4

Solution

  • mine
    • Java
      • Recursion Runtime: 0 ms, faster than 100.00%, Memory Usage: 39 MB, less than 34.92% of Java online submissions 
        public TreeNode trimBST(TreeNode root, int low, int high) {
    if (root == null) 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, root.val);
    root.right = trimBST(root.right, root.val, high);
    return root;
}