Binary Search Tree 
      
    
      
      
      
        Binary Tree 
      
    
      
      
      
        Tree 
      
    
   
  
    
      
       
     
  
  
    
      
    
    
      
       
     
  
Description 
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  
 
 
Follow up:  Could you solve it with time complexity O(height of tree)?
Solutions 
Solution 1 
Python3 Java C++ Go TypeScript Rust 
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26 # Definition for a binary tree node. 
# class TreeNode: 
#     def __init__(self, val=0, left=None, right=None): 
#         self.val = val 
#         self.left = left 
#         self.right = right 
class   Solution : 
    def   deleteNode ( self ,  root :  Optional [ TreeNode ],  key :  int )  ->  Optional [ TreeNode ]: 
        if  root  is  None : 
            return  None 
        if  root . val  >  key : 
            root . left  =  self . deleteNode ( root . left ,  key ) 
            return  root 
        if  root . val  <  key : 
            root . right  =  self . deleteNode ( root . right ,  key ) 
            return  root 
        if  root . left  is  None : 
            return  root . right 
        if  root . right  is  None : 
            return  root . left 
        node  =  root . right 
        while  node . left : 
            node  =  node . left 
        node . left  =  root . left 
        root  =  root . right 
        return  root 
 
 
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43 /** 
 * Definition for a binary tree node. 
 * public class TreeNode { 
 *     int val; 
 *     TreeNode left; 
 *     TreeNode right; 
 *     TreeNode() {} 
 *     TreeNode(int val) { this.val = val; } 
 *     TreeNode(int val, TreeNode left, TreeNode right) { 
 *         this.val = val; 
 *         this.left = left; 
 *         this.right = right; 
 *     } 
 * } 
 */ 
class  Solution   { 
     public   TreeNode   deleteNode ( TreeNode   root ,   int   key )   { 
         if   ( root   ==   null )   { 
             return   null ; 
         } 
         if   ( root . val   >   key )   { 
             root . left   =   deleteNode ( root . left ,   key ); 
             return   root ; 
         } 
         if   ( root . val   <   key )   { 
             root . right   =   deleteNode ( root . right ,   key ); 
             return   root ; 
         } 
         if   ( root . left   ==   null )   { 
             return   root . right ; 
         } 
         if   ( root . right   ==   null )   { 
             return   root . left ; 
         } 
         TreeNode   node   =   root . right ; 
         while   ( node . left   !=   null )   { 
             node   =   node . left ; 
         } 
         node . left   =   root . left ; 
         root   =   root . right ; 
         return   root ; 
     } 
} 
 
 
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32 /** 
 * 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   root ; 
         if   ( root -> val   >   key )   { 
             root -> left   =   deleteNode ( root -> left ,   key ); 
             return   root ; 
         } 
         if   ( root -> val   <   key )   { 
             root -> right   =   deleteNode ( root -> right ,   key ); 
             return   root ; 
         } 
         if   ( ! root -> left )   return   root -> right ; 
         if   ( ! root -> right )   return   root -> left ; 
         TreeNode *   node   =   root -> right ; 
         while   ( node -> left )   node   =   node -> left ; 
         node -> left   =   root -> left ; 
         root   =   root -> right ; 
         return   root ; 
     } 
}; 
 
 
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34 /** 
 * Definition for a binary tree node. 
 * type TreeNode struct { 
 *     Val int 
 *     Left *TreeNode 
 *     Right *TreeNode 
 * } 
 */ 
func   deleteNode ( root   * TreeNode ,   key   int )   * TreeNode   { 
     if   root   ==   nil   { 
         return   nil 
     } 
     if   root . Val   >   key   { 
         root . Left   =   deleteNode ( root . Left ,   key ) 
         return   root 
     } 
     if   root . Val   <   key   { 
         root . Right   =   deleteNode ( root . Right ,   key ) 
         return   root 
     } 
     if   root . Left   ==   nil   { 
         return   root . Right 
     } 
     if   root . Right   ==   nil   { 
         return   root . Left 
     } 
     node   :=   root . Right 
     for   node . Left   !=   nil   { 
         node   =   node . Left 
     } 
     node . Left   =   root . Left 
     root   =   root . Right 
     return   root 
} 
 
 
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45 /** 
 * Definition for a binary tree node. 
 * class TreeNode { 
 *     val: number 
 *     left: TreeNode | null 
 *     right: TreeNode | null 
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) { 
 *         this.val = (val===undefined ? 0 : val) 
 *         this.left = (left===undefined ? null : left) 
 *         this.right = (right===undefined ? null : right) 
 *     } 
 * } 
 */ 
function   deleteNode ( root :   TreeNode   |   null ,   key :   number ) :   TreeNode   |   null   { 
     if   ( root   ==   null )   { 
         return   root ; 
     } 
     const   {   val ,   left ,   right   }   =   root ; 
     if   ( val   >   key )   { 
         root . left   =   deleteNode ( left ,   key ); 
     }   else   if   ( val   <   key )   { 
         root . right   =   deleteNode ( right ,   key ); 
     }   else   { 
         if   ( left   ==   null   &&   right   ==   null )   { 
             root   =   null ; 
         }   else   if   ( left   ==   null   ||   right   ==   null )   { 
             root   =   left   ||   right ; 
         }   else   { 
             if   ( right . left   ==   null )   { 
                 right . left   =   left ; 
                 root   =   right ; 
             }   else   { 
                 let   minPreNode   =   right ; 
                 while   ( minPreNode . left . left   !=   null )   { 
                     minPreNode   =   minPreNode . left ; 
                 } 
                 const   minVal   =   minPreNode . left . val ; 
                 root . val   =   minVal ; 
                 minPreNode . left   =   deleteNode ( minPreNode . left ,   minVal ); 
             } 
         } 
     } 
     return   root ; 
} 
 
 
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71 // Definition for a binary tree node. 
// #[derive(Debug, PartialEq, Eq)] 
// pub struct TreeNode { 
//   pub val: i32, 
//   pub left: Option<Rc<RefCell<TreeNode>>>, 
//   pub right: Option<Rc<RefCell<TreeNode>>>, 
// } 
// 
// impl TreeNode { 
//   #[inline] 
//   pub fn new(val: i32) -> Self { 
//     TreeNode { 
//       val, 
//       left: None, 
//       right: None 
//     } 
//   } 
// } 
use   std :: cell :: RefCell ; 
use   std :: rc :: Rc ; 
impl   Solution   { 
     fn   dfs ( root :   & Option < Rc < RefCell < TreeNode >>> )   ->   i32   { 
         let   node   =   root . as_ref (). unwrap (). borrow (); 
         if   node . left . is_none ()   { 
             return   node . val ; 
         } 
         Self :: dfs ( & node . left ) 
     } 
     pub   fn   delete_node ( 
         mut   root :   Option < Rc < RefCell < TreeNode >>> , 
         key :   i32 , 
     )   ->   Option < Rc < RefCell < TreeNode >>>   { 
         if   root . is_some ()   { 
             let   mut   node   =   root . as_mut (). unwrap (). borrow_mut (); 
             match   node . val . cmp ( & key )   { 
                 std :: cmp :: Ordering :: Less   =>   { 
                     node . right   =   Self :: delete_node ( node . right . take (),   key ); 
                 } 
                 std :: cmp :: Ordering :: Greater   =>   { 
                     node . left   =   Self :: delete_node ( node . left . take (),   key ); 
                 } 
                 std :: cmp :: Ordering :: Equal   =>   { 
                     match   ( node . left . is_some (),   node . right . is_some ())   { 
                         ( false ,   false )   =>   { 
                             return   None ; 
                         } 
                         ( true ,   false )   =>   { 
                             return   node . left . take (); 
                         } 
                         ( false ,   true )   =>   { 
                             return   node . right . take (); 
                         } 
                         ( true ,   true )   =>   { 
                             if   node . right . as_ref (). unwrap (). borrow (). left . is_none ()   { 
                                 let   mut   r   =   node . right . take (); 
                                 r . as_mut (). unwrap (). borrow_mut (). left   =   node . left . take (); 
                                 return   r ; 
                             }   else   { 
                                 let   val   =   Self :: dfs ( & node . right ); 
                                 node . val   =   val ; 
                                 node . right   =   Self :: delete_node ( node . right . take (),   val ); 
                             } 
                         } 
                     }; 
                 } 
             } 
         } 
         root 
     } 
} 
 
 
 
 
  
  
  
    
    
    
    
      
  
    
      
  
     
   
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