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
ac
/**
* 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) {
// exit
if (root == null) return root;
if (key < root.val) {
root.left = deleteNode(root.left, key);
} else if (root.val < key) {
root.right = deleteNode(root.right, key);
} 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 = minValue(root.right);
root.right = deleteNode(root.right, root.val);
}
}
return root;
}
private int minValue(TreeNode root) {
while (root.left != null) {
root = root.left;
}
return root.val;
}
}
/*
find that value: 1) return null if left/right null; 2) return left if right null; 3) return right if left null; 4) find minValue in right, upate in root.val, delete minValue in right, return root.
*/
