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  • 450. 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:

    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

    Recursively find the node that has the same value as the key, while setting the left/right nodes equal to the returned subtree
    Once the node is found, have to handle the below 4 cases

    node doesn't have left or right - return null
    node only has left subtree- return the left subtree
    node only has right subtree- return the right subtree
    node has both left and right - find the minimum value in the right subtree, set that value to the currently found node, then recursively delete the minimum value in the right subtree

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
     /**
      * Recursively find the node that has the same value as the key, while setting the left/right nodes equal to the returned subtree
        Once the node is found, have to handle the below 4 cases
        
        node doesn't have left or right - return null
        node only has left subtree- return the left subtree
        node only has right subtree- return the right subtree
        node has both left and right - find the minimum value in the right subtree, set that value to the currently found node, then recursively delete the minimum value in the right subtree
      */ 
    public class Solution {
        public TreeNode deleteNode(TreeNode root, int 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{  //equal
                if(root.left == null &&  root.right == null) return null;
                else if(root.left == null) return root.right;
                else if(root.right == null) return root.left;
                else{
                    TreeNode minRight = getMin(root.right);
                    root.val = minRight.val;
                    root.right = deleteNode(root.right, minRight.val);
                }
                
            }
            return root;
        }
        public TreeNode getMin(TreeNode root){
            while(root.left != null){
                root = root.left;
            }
            return root;
        }
    }
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  • 原文地址:https://www.cnblogs.com/joannacode/p/6133378.html
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