LeetCode "450. Delete Node in a BST"

Yes the most concise solution is recursion.
https://discuss.leetcode.com/topic/66165/very-concise-c-solution-for-general-binary-tree-not-only-bst

Here is my non-recursive solution - all cases should be covered.

class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if(!root) return nullptr;
        
        bool bIsRoot = root->val == key;
        
        // 1. Find it
        TreeNode *p = root, *pp = nullptr;
        while(p && p->val != key)
        {
            if(p->val == key) break;
            
            pp = p;
            if(p->val < key) p = p->right;
            else p = p->left;
        }
        if(!p) return root; // not found
        
        bool isLeft = pp && pp->left == p;
        
        // 2. Delete it
        if(!p->left && !p->right) // leaf
        {
            if(bIsRoot) return nullptr;
            
            if (isLeft) pp->left = nullptr;
            else pp->right = nullptr;
            
            return root;
        }
        else if(!p->left && p->right) // only right
        {
            if(bIsRoot) return root->right;
            
            if(isLeft) pp->left = p->right;
            else pp->right = p->right;
            
            return root;
        }
        else if(p->left && !p->right)
        {
            if(bIsRoot) return root->left;
            
            if(isLeft) pp->left = p->left;
            else pp->right = p->left;
            
            return root;
        }
        else // 
        {
            TreeNode *q = p->right;
            while(q->left) q = q->left;
            q->left = p->left;
                
            if(bIsRoot)
            {
                return root->right;
            }
            else
            {
                if(isLeft) pp->left = p->right;
                else pp->right = p->right;
                
                return root;
            }
        }
        
        return nullptr;
    }
};