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: Search for a node to remove.
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
recursively find the node that needs to be deleted
Once the node is found, have to handle the below 4 cases
- node doesn't have left nor right - return null
- node only has left subtree- return the left subtree
- node only has right subtree- return the right subtree
- node has both left and right - find the minimum value in the right subtree, set that value to the currently found node, then recursively delete the minimum value in the right subtree
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if (root == null) return null;
if (key < root.val) {
root.left = deleteNode(root.left, key);
}
else if (key > root.val) {
root.right = deleteNode(root.right, key);
}
else { //k == root.val
if (root.left == null) return root.right;
else if (root.right == null) return root.left;
else { // both root.left and root.right are not null
TreeNode minRight = findMin(root.right);
root.val = minRight.val;
root.right = deleteNode(root.right, minRight.val);
}
}
return root;
} public TreeNode findMin(TreeNode cur) {
while (cur.left != null) {
cur = cur.left;
}
return cur;
}
}