[LeetCode]Delete Node in a BST

题目描述:

LeetCode 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

题目大意:

给定一个BST的根节点与一个key,删除BST中key对应的节点。返回BST根节点的引用(有可能被更新)。

基本上,删除操作分为两个阶段:

  1. 寻找待删除节点。
  2. 如果节点找到,删掉这个节点。

注意:时间复杂度为O(树的高度)。

解题思路:

首先根据BST(二叉查找树)的性质,找到目标节点cur及其父节点pre

如果cur不存在,则直接返回root

记ncur为取代cur位置的节点,默认令ncur = cur.right

如果cur拥有左孩子,则将其右孩子链接到左孩子的最大子节点的右侧,令cur = cur.left

然后修正pre与ncur之间的关系

Python代码:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def deleteNode(self, root, key):
        """
        :type root: TreeNode
        :type key: int
        :rtype: TreeNode
        """
        pre, cur = None, root
        while cur and cur.val != key:
            pre = cur
            if key < cur.val:
                cur = cur.left
            elif key > cur.val:
                cur = cur.right
        if not cur: return root

        ncur = cur.right
        if cur.left:
            ncur = cur.left
            self.maxChild(cur.left).right = cur.right

        if not pre: return ncur

        if pre.left == cur:
            pre.left = ncur
        else:
            pre.right = ncur
        return root

    def maxChild(self, root):
        while root.right:
            root = root.right
        return root

上述解法可能会增加BST的深度,导致查询效率降低。

下面是一种递归解法,不会增加BST的深度。

参阅LeetCode Discuss:https://discuss.leetcode.com/topic/65792/recursive-easy-to-understand-java-solution

Java代码:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
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{
            if(root.left == null){
                return root.right;
            }else if(root.right == null){
                return root.left;
            }
            root.val = findMax(root.left).val;
            root.left = deleteNode(root.left, root.val);
        }
        return root;
    }
    
    private TreeNode findMax(TreeNode node){
        while(node.right != null){
            node = node.right;
        }
        return node;
    }
}

 

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