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.
Follow up: Can you solve it with time complexity O(height of tree)?
Example 1:
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.
Example 2:
Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.
Example 3:
Input: root = [], key = 0
Output: []
Constraints:
- The number of nodes in the tree is in the range [0, 104].
- -105 <= Node.val <= 105
- Each node has a unique value.
- root is a valid binary search tree.
- -105 <= key <= 105
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode deleteNode(TreeNode root, int key) {
if(root == null){
return null;
}else if(key < root.val){
root.left=deleteNode(root.left,key);
}else if(key > root.val){
root.right=deleteNode(root.right,key);
}else{
// Case 1: When there is no child, we can set the value to null as we have to delete that
// particular value
if(root.left == null && root.right == null){
root = null;
}
// Case 2: When there is one child
else if(root.left == null){
root = root.right;
}
else if(root.right == null){
root = root.left;
}
// Case 3: When there are two children
else {
TreeNode temp = root;
// Finding minimum element from right or
// Find maximum element from left subtree
TreeNode minNodeFromRight = minimumElement(temp.right);
// here we are replacing current node with the minimum node from the right subtree
root.val = minNodeFromRight.val;
// deleting minimum node
root.right = deleteNode(root.right, minNodeFromRight.val);
}
}
return root;
}
// Get minimum element in binary search tree
// we keep going on left as minimum will be on left
public static TreeNode minimumElement(TreeNode root){
if (root.left == null)
return root;
else
return minimumElement(root.left);
}
}- case 1: if node's both left and right are null, just remove the node;
- case 2: node has only left child: make parent.left child = nodes.left;
- case 3: node has only right child: parnt.right = node.right;
- case 4: node has both child: get left most child of right subtree. replace the value of current; remove the leftmost child of right sub tree;