原题链接在这里:
题目:
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 = 2Output: 1 \ 2
Example 2:
Input: 3 / \ 0 4 \ 2 / 1 L = 1 R = 3Output: 3 / 2 / 1
题解:
在[L, R]区间外trim.
Recursive call, stop condition, root == null, return root.
如果root.val 小于 L, 则root右侧有可能在区间内,返回trim后的 root.right.
如果root.val 大于 R, 则root左侧有可能在区间内,返回trim后的root.left.
如果root.val在区间内, 则对左右两侧分别trim后return root.
Time Complexity: O(n), n 是node数.
Space: O(logn), stack space.
AC Java:
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */10 class Solution {11 public TreeNode trimBST(TreeNode root, int L, int R) {12 if(root == null){13 return root;14 }15 16 if(root.val < L){17 return trimBST(root.right, L, R);18 }19 if(root.val > R){20 return trimBST(root.left, L, R);21 }22 23 root.left = trimBST(root.left, L, R);24 root.right = trimBST(root.right, L, R);25 26 return root;27 }28 }