#450 Leetcode Delete Node in a BST Solution in C, C++, Java, JavaScript, Python, C# Leetcode

Algorithm

Problem Name: 450. Delete Node in a BST

Problem Link: https://leetcode.com/problems/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.

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].
-10⁵ <= Node.val <= 10⁵
Each node has a unique value.
root is a valid binary search tree.
-10⁵ <= key <= 10⁵

Code Examples

#1 Code Example with C Programming

Code - C Programming


struct TreeNode* deleteNode(struct TreeNode* root, int key) {
    struct TreeNode *p, **pp;
    struct TreeNode *a, *b;
    pp = &root;
    p = *pp;
    while (p) {
        if (p->val > key) {
            pp = &p->left;
        } else if (p->val  <  key) {
            pp = &p->right;
        } else {
            break;
        }
        p = *pp;
    }
    if (p) {
        // found it!
        if (!p->right) {
            // free(p);
            *pp = p->left;
        } else if (!p->left) {
            // free(p);
            *pp = p->right;
        } else {
            a = p->left;
            b = p->right;
            // free(p);
            *pp = a;
            while (a->right) {
                a = a->right;
            }
            a->right = b;
        }
    }
    return root;
}

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Input

–

cmd

root = [5,3,6,2,4,null,7], key = 3

Output

–

cmd

[5,4,6,2,null,null,7]

#2 Code Example with C++ Programming

Code - C++ Programming


/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        TreeNode* head = new TreeNode(0);
        head->left = root;
        dfs(head, root, key);
        return head->left;
    }
    
    void dfs(TreeNode* pre, TreeNode* cur, int key){
        if(!cur) return;
        cur->val > key ? dfs(cur, cur->left, key) : cur->val  <  key ? dfs(cur, cur->right, key) : helper(pre, cur);
    }
    
    void helper(TreeNode* pre, TreeNode* cur){
        TreeNode* p = cur->left;
        while(p && p->right) p = p->right;
        p ? p->right = cur->right : cur->left = cur->right;
        cur == pre->left ? pre->left = cur->left : pre->right = cur->left;
    }
};


/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) {
            return NULL;
        }
        TreeNode* head = new TreeNode(0);
        head->left = root;
        TreeNode* parent;
        TreeNode* node = dfs(root, head, key, parent);
        if (!node) {
            return root;
        }
        
        TreeNode* l = node->left;
        TreeNode* r = node->right;
        TreeNode* p = r;
        while (p && p->left) {
            p = p->left;
        }
        if (!r) {
            r = l;
        } else if (p) {
            p->left = l;
        }
        if (node == parent->left) {
            parent->left = r;
        } else {
            parent->right = r;
        }
        return head->left;
    }
    
    TreeNode* dfs(TreeNode* cur, TreeNode* pre, int key, TreeNode*& parent) {
        if (!cur) {
            return NULL;
        }
        if (cur->val == key) {
            parent = pre;
            return cur;
        }
        
        if (cur->val > key) {
            return dfs(cur->left, cur, key, parent);
        } else {
            return dfs(cur->right, cur, key, parent);
        }
    }
};

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Input

–

cmd

root = [5,3,6,2,4,null,7], key = 3

Output

–

cmd

[5,4,6,2,null,null,7]

#3 Code Example with Java Programming

Code - Java Programming


class Solution {
  public TreeNode deleteNode(TreeNode root, int key) {
    if (root == null) {
      return null;
    }
    if (key > root.val) {
      root.right = deleteNode(root.right, key);
    } else if (key  <  root.val) {
      root.left = deleteNode(root.left, key);
    } else {
      if (root.left == null || root.right == null) {
        return root.left == null ? root.right : root.left;
      } 
      TreeNode inorderSuccessor = root.right;
      while (inorderSuccessor.left != null) {
        inorderSuccessor = inorderSuccessor.left;
      }
      root.val = inorderSuccessor.val;
      root.right = deleteNode(root.right, inorderSuccessor.val);
    }
    return root;
  }
}

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Input

–

cmd

root = [5,3,6,2,4,null,7], key = 0

Output

–

cmd

[5,3,6,2,4,null,7]

#4 Code Example with Javascript Programming

Code - Javascript Programming


const deleteNode = function(root, key) {
  if(root == null) return null
  if(key < root.val) {
    root.left = deleteNode(root.left, key>
  } else if(key > root.val) {
    root.right = deleteNode(root.right, key)
  } else {
    if(root.left == null) {
      return root.right
    } else if(root.right == null) {
      return root.left
    } else {
      let smallestRight = root.right
      while(smallestRight.left !== null) smallestRight = smallestRight.left
      smallestRight.left = root.left
      return root.right
    }
  }
  
  return root
};

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Input

–

cmd

root = [5,3,6,2,4,null,7], key = 0

Output

–

cmd

[5,3,6,2,4,null,7]

#5 Code Example with Python Programming

Code - Python Programming


class Solution:
    def deleteNode(self, root, key):
        """
        :type root: TreeNode
        :type key: int
        :rtype: TreeNode
        """
        if not root: return
        if root.val > key: root.left = self.deleteNode(root.left, key)
        elif root.val < key: root.right = self.deleteNode(root.right, key)
        else:
            if not root.right: return root.left
            elif not root.left: return root.right
            tmp, mini = root.right, root.right.val
            while tmp.left:
                tmp, mini = tmp.left, tmp.left.val
            root.val, root.right = mini, self.deleteNode(root.right, mini)
        return root

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Input

–

cmd

root = [], key = 0

Output

–

cmd

[]

#6 Code Example with C# Programming

Code - C# Programming


namespace LeetCode
{
    public class _0450_DeleteNodeInABST
    {
        public TreeNode DeleteNode(TreeNode root, int key)
        {
            if (root == null) return null;

            if (key > root.val) root.right = DeleteNode(root.right, key);
            else if (key  <  root.val) root.left = DeleteNode(root.left, key);
            else
            {
                if (root.left == null && root.right == null)
                    return null;
                else if (root.right != null)
                {
                    root.val = GetSuccessor(root);
                    root.right = DeleteNode(root.right, root.val);
                }
                else
                {
                    root.val = GetPredecessor(root);
                    root.left = DeleteNode(root.left, root.val);
                }
            }

            return root;
        }

        private int GetSuccessor(TreeNode node)
        {
            node = node.right;
            while (node.left != null)
                node = node.left;

            return node.val;
        }

        private int GetPredecessor(TreeNode node)
        {
            node = node.left;
            while (node.right != null)
                node = node.right;

            return node.val;
        }
    }
}

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Input

–

cmd

root = [], key = 0

Output

–

cmd

[]

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