450. Delete Node in a BST

Medium

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.

Example 1:

Input: root = [5,3,6,2,4,null,7], key = 3
Output:
 [5,4,6,2,null,null,7]
Explanation:
 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 above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.

Example 2:

Input: root = [5,3,6,2,4,null,7], key = 0
Output:
 [5,3,6,2,4,null,7]
Explanation:
 The tree does not contain a node with value = 0.

Example 3:

Input: root = [], key = 0
Output:
 []

Constraints:

The number of nodes in the tree is in the range [0, 104].
-105 <= Node.val <= 105
Each node has a unique value.
root is a valid binary search tree.
-105 <= key <= 105

Follow up: Could you solve it with time complexity O(height of tree)?

解題

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func deleteNode(root *TreeNode, key int) *TreeNode {
    if root == nil {
        return root
    }

    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 == nil && root.Right == nil { //為left node ，直接回傳nil即可
            return nil
        } 

        if root.Left == nil { // 左子樹為空，回傳右子樹
            temp := root.Right
            root.Right = nil
            return temp
        }

        if root.Right == nil { //右子樹為空，回傳左子樹
            temp := root.Left
            root.Left = nil
            return temp
        }

        temp := succesor(root.Right) //從右邊選出最小的值作為繼任的節點
        root.Val = temp.Val
        root.Right = deleteNode(root.Right, temp.Val) //將繼任的節點從右邊刪除
    }

    return root
}

func succesor(root *TreeNode) *TreeNode {
    for root.Left != nil {
        root = root.Left
    }
    return root
}

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Last updated 2 years ago

Input: root = [5,3,6,2,4,null,7], key = 3 Output: [5,4,6,2,null,null,7] Explanation: 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 above BST. Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.

/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func deleteNode(root *TreeNode, key int) *TreeNode { if root == nil { return root } 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 == nil && root.Right == nil { //為left node ，直接回傳nil即可 return nil } if root.Left == nil { // 左子樹為空，回傳右子樹 temp := root.Right root.Right = nil return temp } if root.Right == nil { //右子樹為空，回傳左子樹 temp := root.Left root.Left = nil return temp } temp := succesor(root.Right) //從右邊選出最小的值作為繼任的節點 root.Val = temp.Val root.Right = deleteNode(root.Right, temp.Val) //將繼任的節點從右邊刪除 } return root } func succesor(root *TreeNode) *TreeNode { for root.Left != nil { root = root.Left } return root }