Leetcode: Delete Node in a BST

Delete Node in a BST

Similar Problems:

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

  1. Search for a node to remove.
  2. If the node is found, delete the node.

Note: Time complexity should be O(height of tree).


root = [5,3,6,2,4,null,7]
key = 3

   / \
  3   6
 / \   \
2   4   7

Given key to delete is 3. So we find the node with value 3 and delete it.

One valid answer is [5,4,6,2,null,null,7], shown in the following BST.

   / \
  4   6
 /     \
2       7

Another valid answer is [5,2,6,null,4,null,7].

   / \
  2   6
   \   \
    4   7

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

## Blog link: https://code.dennyzhang.com/delete-node-in-a-bst
## Basic Ideas: Find the target
##              If the target is a leaf, we need the parent node
##              If the target only have one child
##              If the target has both children
## Complexity: Time O(h), Space O(1)
# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def deleteNode(self, root, key):
        :type root: TreeNode
        :type key: int
        :rtype: TreeNode
        if root is None:
            return None

        # delete root
        if root.val == key:
            # root is a leaf
            if root.left is None and root.right is None:
                return None
                self.deleteRootNode(root, None)
                return root

        # delete node inside the tree
        prev, node = root, root
        while node:
            if node.val == key:
            prev = node
            if node.val > key:
                node = node.left
                node = node.right

        # node is the target to be deleted
        if node:
            self.deleteRootNode(node, prev)
        return root

    def deleteRootNode(self, root, prev):
        if root is None:
        # leaf
        if root.left is None and root.right is None:
            if root == prev.left:
                prev.left = None
                prev.right = None
        if root.left:
            # find the largest item in the left right
            p, node = root, root.left
            while node.right:
                p = node
                node = node.right
            # swap value
            root.val = node.val
            return self.deleteRootNode(node, p)
        if root.right:
            p, node = root, root.right
            while node.left:
                p = node
                node = node.left
            root.val = node.val
            return self.deleteRootNode(node, p)

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