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
/**
* 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 val) {
if (root == null) return root; if (root.val > val) {
root.left = deleteNode(root.left, val);
} else if (root.val < val) {
root.right = deleteNode(root.right, val);
} else {
if (root.left == null && root.right == null) {
return null;
} else if (root.left == null) {
return root.right;
} else if (root.right == null) {
return root.left;
} else {
root.val = findMin(root.right).val;
root.right = deleteNode(root.right, root.val);
}
}
return root;
} public TreeNode findMin(TreeNode n) {
if (n.left != null) {
return findMin(n.left);
}
return n;
}
}