标签：binary val new else nbsp hang element 移除 out
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
正确方法其实应该是在遍历的过程中就修改二叉树，移除不合题意的结点。当然对于二叉树的题，十有八九都是要用递归来解的。首先判断如果root为空，那么直接返回空即可。然后就是要看根结点是否在范围内，如果根结点值小于L，那么返回对其右子结点调用递归函数的值；如果根结点大于R，那么返回对其左子结点调用递归函数的值。如果根结点在范围内，将其左子结点更新为对其左子结点调用递归函数的返回值，同样，将其右子结点更新为对其右子结点调用递归函数的返回值。最后返回root即可，参见代码如下：
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public TreeNode trimBST(TreeNode root, int L, int R) { if(root == null){ return null; } if(root.val > R){ return trimBST(root.left, L, R); } else if(root.val < L){ return trimBST(root.right, L, R); } root.left = trimBST(root.left, L, R); root.right = trimBST(root.right, L, R); return root; } }
LeetCode - Trim a Binary Search Tree
标签：binary val new else nbsp hang element 移除 out
原文地址：https://www.cnblogs.com/incrediblechangshuo/p/9696314.html
