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l450 delete node in a bst

admin 11月 13, 2020 0

题目描述

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

二分遍历BST，发现对应的节点root，查找右子树最小的元素current，将root.left接到current.left上，返回root.right

Go代码实现

Go代码实现1

func (root *TreeNode, key int) *TreeNode {
    if root == nil{
        return root
    }

    root = findDelete(root, key)
    return root
}

func findDelete(root *TreeNode, key int)*TreeNode{
    if root == nil {
        return root
    }

    if root.Val == key{
        if root.Left == nil{
            return root.Right
        }else if root.Right == nil {
            return root.Left
        }

        current := root.Right
        for current.Left != nil {
            current = current.Left
        }
        current.Left = root.Left
        return root.Right

    }else if root.Val>key {
        root.Left = findDelete(root.Left, key)
    }else{
        root.Right = findDelete(root.Right, key)
    }
    return root
}

Runtime: 1162 ms 27.78%

参考

https://leetcode.com/problems/delete-node-in-a-bst/discuss/93296/