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.
Note: Time complexity should be O(height of tree).

Solution：

因为是BST，所以先利用二分查找到要删除的node，若不是BST则需要依次找(Solution2)，找到node后的要删除的几种情况：
node doesn't have left or right - return null
node only has left subtree- return the left subtree
node only has right subtree- return the right subtree

• node has both left and right - find the minimum value in the right subtree, 就是右树的最左下node_lb，将其替换要删除的node(替换连接)，并按照node_lb only has right subtree的方式删除node_lb
  Time Complexity: O(N) Space Complexity: O(1) (worst case O(n) for 递归缓存)

  
  
  IMG_0143.JPG

Solution1 Code：

class Solution {
    private TreeNode deleteRootNode(TreeNode root) {
        if (root == null) {
            return null;
        }
        if (root.left == null) {
            return root.right;
        }
        if (root.right == null) {
            return root.left;
        }
        TreeNode next = root.right;
        TreeNode pre = null;
        for(; next.left != null; pre = next, next = next.left);
        next.left = root.left;
        if(root.right != next) {
            pre.left = next.right;
            next.right = root.right;
        }
        return next;
    }
    
    public TreeNode deleteNode(TreeNode root, int key) {
        TreeNode cur = root;
        TreeNode pre = null;
        while(cur != null && cur.val != key) {
            pre = cur;
            if (key < cur.val) {
                cur = cur.left;
            } else if (key > cur.val) {
                cur = cur.right;
            }
        }
        if (pre == null) {
            return deleteRootNode(cur);
        }
        if (pre.left == cur) {
            pre.left = deleteRootNode(cur);
        } else {
            pre.right = deleteRootNode(cur);
        }
        return root;
    }
}

**Solution2 Code: **

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return nullptr;
        if (root->val == key) {
            if (!root->right) {
                TreeNode* left = root->left;
                delete root;
                return left;
            }
            else {
                TreeNode* right = root->right;
                while (right->left)
                    right = right->left;
                swap(root->val, right->val);    
            }
        }
        root->left = deleteNode(root->left, key);
        root->right = deleteNode(root->right, key);
        return root;
    }
};