450. Delete Node in a BST

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LeetCode

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Problem

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].
• -10^5 <= Node.val <= 10^5
• Each node has a unique value.
• root is a valid binary search tree.
• -10^5 <= key <= 10^5

Code

669 705

class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null) return root;

        // 三种情况
        if(key > root.val){
            root.right = deleteNode(root.right, key);
        } else if (key < root.val){
            root.left = deleteNode(root.left, key);
        // 需要删除root
        } else {
            // 三种情况
            if(root.left == null) {
                return root.right;
            } else if (root.right == null) {
                return root.left;
            } else{
                // 找到右子树的最小值
                TreeNode min = root.right;
                while(min.left != null){
                    min = min.left;
                }

                // 把min节点的值赋值给root
                root.val = min.val;
                // 删除右子树中的min节点
                root.right = deleteNode(root.right, min.val);
            }
        }

        return root;
    }
}