Description
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
image.png
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.
image.png
Another valid answer is [5,2,6,null,4,null,7].
image.png
Solution
Recursion, O(h), S(h)
/**
* 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 null;
}
if (root.val < key) {
root.right = deleteNode(root.right, key);
} else if (root.val > key) {
root.left = deleteNode(root.left, key);
} else {
if (root.left == null) {
return root.right;
} else if (root.right == null) {
return root.left;
}
int min = findMin(root.right);
root.val = min;
root.right = deleteNode(root.right, min);
}
return root;
}
private int findMin(TreeNode root) {
while (root != null && root.left != null) {
root = root.left;
}
return root.val;
}
}
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