Formatted question description: https://leetcode.ca/all/450.html

450. Delete Node in a BST (Medium)

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).

Example:

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

    5
   / \
  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.

    5
   / \
  4   6
 /     \
2       7

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

    5
   / \
  2   6
   \   \
    4   7

Companies:
Microsoft, Google, Amazon

Related Topics:
Tree

Solution 1.

// OJ: https://leetcode.com/problems/delete-node-in-a-bst/

// Time: O(H)
// Space: O(H)
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return NULL;
        if (root->val > key) root->left = deleteNode(root->left, key);
        else if (root->val < key) root->right = deleteNode(root->right, key);
        else if (root->left) {
            auto p = root->left;
            while (p->right) p = p->right;
            root->val = p->val;
            root->left = deleteNode(root->left, root->val);
        } else if (root->right) {
            auto p = root->right;
            while (p->left) p = p->left;
            root->val = p->val;
            root->right = deleteNode(root->right, root->val);
        } else {
            delete root;
            root = NULL;
        }
        return root;
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/delete-node-in-a-bst/

// Time: O(H)
// Space: O(H)
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return NULL;
        if (root->val > key) root->left = deleteNode(root->left, key);
        else if (root->val < key) root->right = deleteNode(root->right, key);
        else if (!root->left) {
            auto right = root->right;
            delete root;
            return right;
        } else if (!root->right) {
            auto left = root->left;
            delete root;
            return left;
        } else {
            auto node = root->right;
            while (node->left) node = node->left;
            root->val = node->val;
            root->right = deleteNode(root->right, root->val);
        }
        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 deleteNode(TreeNode root, int key) {
        if (root == null)
            return root;
        if (root.val == key) {
            TreeNode dummyRoot = new TreeNode(0);
            dummyRoot.left = root;
            if (root.right != null) {
                TreeNode successor = root.right;
                if (successor.left == null) {
                    successor.left = root.left;
                    dummyRoot.left = successor;
                } else {
                    TreeNode parent = successor, child = successor.left;
                    while (child.left != null) {
                        child = child.left;
                        parent = parent.left;
                    }
                    root.val = child.val;
                    parent.left = child.right;
                }
            } else
                dummyRoot.left = root.left;
            return dummyRoot.left;
        } else {
            TreeNode parent = root;
            TreeNode child = root.val > key ? root.left : root.right;
            while (child != null && child.val != key) {
                parent = child;
                child = child.val > key ? child.left : child.right;
            }
            if (child != null) {
                boolean isLeftChild = parent.left == child;
                if (child.left == null && child.right == null) {
                    if (isLeftChild)
                        parent.left = null;
                    else
                        parent.right = null;
                } else if (child.left == null) {
                    if (isLeftChild)
                        parent.left = child.right;
                    else
                        parent.right = child.right;
                } else if (child.right == null) {
                    if (isLeftChild)
                        parent.left = child.left;
                    else
                        parent.right = child.left;
                } else {
                    TreeNode temp1 = child.right;
                    if (temp1.left == null) {
                        child.val = temp1.val;
                        child.right = temp1.right;
                    } else {
                        TreeNode temp2 = temp1.left;
                        while (temp2.left != null) {
                            temp2 = temp2.left;
                            temp1 = temp1.left;
                        }
                        child.val = temp2.val;
                        temp1.left = temp2.right;
                    }
                }
            }
            return root;
        }
    }
}

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