0450.md

450. Delete Node in a BST

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Topic:

Question

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.

Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: 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 above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted. \

Intuition

Basically, the deletion can be divided into two stages:

1. Search for a node to remove. O(log n)
2. If the node is found, delete the node. O(1)

Delete the node:

• Balance the tree by moving the right child up
• assign left child to right child's leftmost (left child must be smaller than right's leftmost)

COULD BE IMPROVED

Code

Time: O(log n)
Space: O(h)

Recursive

public TreeNode deleteNode(TreeNode root, int key) {
    TreeNode dummy = new TreeNode();
    dummy.right = root;
    TreeNode parent = dummy;

    parent = findDeletedParent(root, parent, key);
    if (parent != null) {
        boolean isLeftChild = parent.left != null && parent.left.val == key;
        TreeNode deleted = isLeftChild ? parent.left:parent.right;
        TreeNode left = deleted.left;
        TreeNode right = deleted.right;

        // balance the tree
        // move the right subtree up
        if (isLeftChild)
            parent.left = right;
        else
            parent.right = right;

        // put left subtree on the right subtree's leftmost
        if (right != null) {
            while (right.left != null)
                right = right.left;
            right.left = left;
        }
        // if the right subtree is empty, move the left subtree up
        else {
            if (isLeftChild)
                parent.left = left;
            else
                parent.right = left;
        }
    }
    return dummy.right;
}

private TreeNode findDeletedParent(TreeNode root, TreeNode parent, int key) {
    if (root == null) return null;

    if (root.val == key) return parent;

    TreeNode left = null, right = null;
    if (root.val < key) right = findDeletedParent(root.right, root, key);
    if (root.val > key) left = findDeletedParent(root.left, root, key);

    return left == null ? right:left;
}