450. Delete Node in a BST

---

My Solution

Solution 1: DFS (Iteration by using BST property)

The Key Idea for Solving This Coding Question

C++ Code

/**
 * 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) {
        TreeNode *delNode = root, *parent = nullptr;
        while (delNode != nullptr) {
            if (delNode->val < key) {
                parent = delNode;
                delNode = delNode->right;
            } else if (key < delNode->val) {
                parent = delNode;
                delNode = delNode->left;
            } else {
                break;  // find key
            }
        }
        if (delNode == nullptr) {
            return root; // key doesn't exist in BST.
        }

        if (parent == nullptr) {
            // root->val == key is true. Delete root.
            return deleteTreeNode(root);
        } else if (parent->left != nullptr && parent->left == delNode) {
            parent->left = deleteTreeNode(delNode);
        } else {
            parent->right = deleteTreeNode(delNode);
        }
        return root;
    }

private:
    TreeNode *deleteTreeNode(TreeNode *delNode) {
        if (delNode->left == nullptr) {
            TreeNode *x = delNode->right;
            delete delNode;
            return x;
        }

        TreeNode *predecessor = delNode->left;
        while (predecessor->right != nullptr) {
            predecessor = predecessor->right;
        }

        TreeNode *y = delNode->left;
        predecessor->right = delNode->right;
        delete delNode;
        return y;
    }
};

Time Complexity

$O(H)$
H is the height of the BST.

Space Complexity

$O(1)$

Solution 2: DFS (Recursion by using BST property)

The Key Idea for Solving This Coding Question

C++ Code 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 == nullptr) {
            return nullptr;
        }

        if (root->val < key) {
            root->right = deleteNode(root->right, key);
            return root;
        }
        if (key < root->val) {
            root->left = deleteNode(root->left, key);
            return root;
        }

        // root->val == key is true. root should be deleted.
        if (root->left != nullptr) {
            int predVal = findPredVal(root);
            root->val = predVal;
            root->left = deleteNode(root->left, predVal);
            return root;
        }
        if (root->right != nullptr) {
            int succVal = findSuccVal(root);
            root->val = succVal;
            root->right = deleteNode(root->right, succVal);
            return root;
        }

        // root is a leaf
        delete root;
        return nullptr;
    }

private:
    int findPredVal(TreeNode *root) {
        TreeNode *predecessor = root->left;
        while (predecessor->right != nullptr) {
            predecessor = predecessor->right;
        }

        return predecessor->val;
    }

    int findSuccVal(TreeNode *root) {
        TreeNode *successor = root->right;
        while (successor->left != nullptr) {
            successor = successor->left;
        }

        return successor->val;
    }
};

C++ Code 2

/**
 * 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 == nullptr) {
            return nullptr;
        }
        
        if (root->val < key) {
            root->right = deleteNode(root->right, key);
        } else if (key < root->val) {
            root->left = deleteNode(root->left, key);
        } else {
            // root->val == key is true.
            if (root->left == nullptr && root->right == nullptr) {
                delete root;
                return nullptr;
            }
            
            if (root->left == nullptr) {
                TreeNode *x = root->right;
                delete root;
                return x;
            }
            
            if (root->right == nullptr) {
                TreeNode *y = root->left;
                delete root;
                return y;
            }
            
            int predVal = findPredecessorVal(root);
            root->val = predVal;
            root->left = deleteNode(root->left, predVal);
        }
        
        return root;

    }
private:
    int findPredecessorVal(TreeNode *node) {
        TreeNode *predecessor = node->left;
        while (predecessor->right != nullptr) {
            predecessor = predecessor->right;
        }
        return predecessor->val;
    }

};

Time Complexity

$O(H)$
H is the height of the BST

Space Complexity

$O(H)$
H is the height of the BST

Miscellane