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DeleteNodeInBST.java
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DeleteNodeInBST.java
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package binary_search_tree;
import utils.TreeNode;
/**
* 450. Delete Node in a BST
*
* 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.
*
* IMP-2 : This problem is important to understand. Once logic is understood , its fairly simple
*/
public class DeleteNodeInBST {
/**
* this is a tricky problem but can be simplified via recursion
* @param root
* @param key
* @return
*/
TreeNode deleteNode(TreeNode root, int key) {
if(root == null) return null; //edge case
if(root.val == key) { //here is the case where values match
//case where left or right or both are null, return the non-null child
if (root.left == null) {
return root.right;
}
if (root.right == null) {
return root.left;
}
//in case where both left and right are not null recurse down to the in order successor
//delete that node and swap its values with the node to be deleted.
TreeNode next = root.right;
while (next.left != null) {
next = next.left;
}
deleteNode(root, next.val);
root.val = next.val;
return root;
} else if (root.val > key && root.left != null) {
//set the return value to left node so as to set tree structure correctly.
root.left = deleteNode(root.left, key);
} else if (root.val < key && root.right != null) {
root.right = deleteNode(root.right, key);
}
return root;
}
}