Given a binary search tree and the lowest and highest boundaries as L
and R
, trim the tree so that all its elements lies in [L, R]
(R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.
Example 1:
Input: 1 / 0 2 L = 1 R = 2 Output: 1 2
Example 2:
Input: 3 / 0 4 2 / 1 L = 1 R = 3 Output: 3 / 2 / 1
思路:
BST的特性:left < root < right
case 1: 如果root.val < L, 则返回修剪后的右子树 (把root 和左子树都裁掉:左子树的所有值一定小于L)
case 2: 如果root.val > R, 则返回修剪后的左子树 (裁掉root 和 右子树)
case 3: 如果 root.val 在范围内,那么保留root结点本身, 并且root.left = 修剪后的左子树, root.right = 修剪后的右子树。
代码:
/** * 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 L, int R) { if (root == null) return root; if (root.val < L) return trimBST(root.right, L, R); if (root.val > R) return trimBST(root.left, L, R); // 要把修剪过后的左右子树赋值给root.left root.right root.left = trimBST(root.left, L, R); root.right = trimBST(root.right, L, R); return root; } }