450. Delete Node in a BST
by Botao Xiao
450. Delete Node in a BST
Question
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
Solution
- Method 1: recursion
```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(key < root.val) root.left = deleteNode(root.left, key); else if(key > root.val) root.right = deleteNode(root.right, key); else{ TreeNode res = null; if(root.left == null) res = root.right; else if(root.right == null) res = root.left; else{ // both left and right are not null. TreeNode parent = root; res = root.right; while(res.left != null){ parent = res; res = res.left; } if(parent != root){ parent.left = res.right; res.right = root.right; } res.left = root.left; } return res; } return root; } } ```
Reference
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