LeetCode 669. Trim a Binary Search Tree(修剪二叉搜索树)
题目描述
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
解法一
思路
仔细想想这题其实就是“二叉搜索树中删除一个结点”的变形,所以可以直接套用那个模板了:
def trimBST(self, root, L, R):
if not root:
return root
# TODO
root.left = self.trimBST(root.left, L, R)
root.right = self.trimBST(root.right, L, R)
return root
那么下面的问题就是TODO
该怎么写了。再来看看题目,如果当前结点的值不在[L, R]
之间,那么这个结点应该删除掉,那么以哪个结点来替换呢?这个倒不难:如果小于L,那么一定是右子树结点;如果大于R,那么一定是左子树结点。注意这里只是说“子树结点”而不是“子节点”,是因为当前结点的“子节点”也可能不满足要求。那么怎么寻找这个“子树结点”呢?我们只用交给trimBST()
来处理啦,这里可能有点难理解,还需自行体会。
所以TODO
部分就是:
if root.val < L:
return self.trimBST(root.right, L, R)
if root.val > R:
return self.trimBST(root.left, L, R)
Python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def trimBST(self, root, L, R):
"""
:type root: TreeNode
:type L: int
:type R: int
:rtype: TreeNode
"""
if not root:
return root
if root.val < L:
return self.trimBST(root.right, L, R)
if root.val > R:
return self.trimBST(root.left, L, R)
root.left = self.trimBST(root.left, L, R)
root.right = self.trimBST(root.right, L, R)
return root
Java
/**
* 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 root;
}
if (root.val < L) {
return trimBST(root.right, L, R);
}
if (root.val > R) {
return trimBST(root.left, L, R);
}
root.left = trimBST(root.left, L, R);
root.right = trimBST(root.right, L, R);
return root;
}
}
C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *trimBST(TreeNode *root, int L, int R) {
if (root == nullptr) {
return root;
}
if (root->val < L) {
return trimBST(root->right, L, R);
}
if (root->val > R) {
return trimBST(root->left, L, R);
}
root->left = trimBST(root->left, L, R);
root->right = trimBST(root->right, L, R);
return root;
}
};