[Leetcode-Tree]Delete Node in a BST

wangjuntytl 发布于2019-08-16 10:26 / 1251人阅读

摘要：解题思路我们可以用递归来查找，在找到需要删除的节点后，我们需要分情况讨论节点是叶子节点，直接返回节点有一个孩子，直接返回孩子节点有两个孩子，我们要将右子树中最小的节点值赋值给根节点，并在右子树中删除掉那个最小的节点。

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.
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

解题思路

我们可以用递归来查找Key，在找到需要删除的节点后，我们需要分情况讨论：
1）节点是叶子节点，直接返回null;
2) 节点有一个孩子，直接返回孩子；
3）节点有两个孩子，我们要将右子树中最小的节点值赋值给根节点，并在右子树中删除掉那个最小的节点。

2.代码

public class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root==null) return root;
        if(key>root.val)
            root.right=deleteNode(root.right,key);
        else if(key

专线服务混合云 BST python bst bst树 DELETE

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