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:

  1. Search for a node to remove.
  2. 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

Time complexity : O(h)
Space complexity : O(1)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return nullptr;
        
        if (key < root->val)
            root->left = deleteNode(root->left, key);
        else if (root->val < key)
            root->right = deleteNode(root->right, key);
        else {
            // root->val == key
            TreeNode* new_node = nullptr;
            if (root->left && root->right) {
                // root has both children
                new_node = root->right;
                TreeNode* parent = root;
                while (new_node->left) {
                    parent = new_node;
                    new_node = new_node->left;
                }
                if (parent != root) {
                    parent->left = new_node->right;
                    new_node->right = root->right;
                }
                
                new_node->left = root->left;
            } else {
                // root has no or one child.
                new_node = root->left ? root->left : root->right;
            }
            delete root;
            return new_node;
        }
        return root;
    }
};

首先要找到目標node，即node->val == key。
因此若key < node->val，代表目標小於當前，則往左子樹走；若node->val < key，代表目標大於當前，則往右子樹走。

找到目標node後，要刪掉它，此時node有以下情況：

  1. 沒有任何child。直接刪掉，return nullptr。
  2. 有左子樹，無右子樹。刪掉後，return 左子樹。
  3. 無左子樹，有右子樹。刪掉後，return 右子樹。
  4. 有左右子樹，此時做法就很多元了，這邊採取的做法是「找到右子樹中值最小的來取代」。

如何「找到右子樹中值最小的node來取代」：
最小的值，意指無左子樹，因此一直往左子樹走，走到無左子樹，就是值最小的node。
找到後，再與目標node銜接即可。