[669] force diary buckle trim binary search tree | recursive

Title Description

Given a binary search tree, while the given minimum and maximum boundary boundary L R. By pruning the binary search tree, so that the values ​​of all the nodes in the [L, R] in the (R> = L). You may need to change the root of the tree, so the results should return good binary search tree pruning the new root node.

Algorithm thinking

Binary Tree Recursive natural fit.

First edition:

class Solution:
    def trimBST(self, root: TreeNode, L: int, R: int) -> TreeNode:
        if root==None:return None
        # if root.val<L:root=root.right
        # elif root.val>R:root=root.left
        # if root==None:return None
        # root.left=self.trimBST(root.left,L,R)
        # root.right=self.trimBST(root.right,L,R)
        # return root

The output is [3,2,4], the answer is [3, null, 4].
Clearly, where a node, root = root.right, then node 2 has not been directly verified

# root.left=self.trimBST(root.left,L,R)
# root.right=self.trimBST(root.right,L,R)

Then returned.

After the simple thought:
Second Edition:

class Solution:
    def trimBST(self, root: TreeNode, L: int, R: int) -> TreeNode:
        if root==None:return None
        if root.val<L:root=self.trimBST(root.right,L,R)
        elif root.val>R:root=self.trimBST(root.left,L,R)
        else:
            root.left=self.trimBST(root.left,L,R)
            root.right=self.trimBST(root.right,L,R)
        return root

When execution: 52 ms, beat the 85.43% of users in all Python3 submission