450-Delete Node in a BST

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:

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).

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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

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問題描述

給定二叉排序樹的根節點和一個鍵， 刪除樹中對應鍵的節點。返回二叉排序樹的根節點(可能會更新)

一般， 刪除可以分為兩部分:

1. 找出刪除的節點
2. 刪除節點

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問題分析

找出刪除節點， 若其左子樹或右子樹為空， 那麼直接返回刪除節點的右子樹或者左子樹， 否則將刪除節點的左子樹作為刪除節點的右子樹的最左節點的左子樹， 返回刪除節點的右子樹的根節點

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解法

class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null) return null;

        if(root.val > key){
            root.left = deleteNode(root.left, key);
        }else if(root.val < key){
            root.right = deleteNode(root.right, key);
        }else{
            if(root.left == null) return root.right;
            else if(root.right == null) return root.left;
            else{
                TreeNode leftMostNodeOfRight = toolFunc(root.right);
                leftMostNodeOfRight.left = root.left;
                return root.right;
            }
        }

        return root;
    }
    public TreeNode toolFunc(TreeNode root){
        while(root.left != null) root = root.left;
        return root;
    }    
}