1. 程式人生 > >Delete Node in a BST 二叉查詢樹的查詢、插入和刪除 - Java實現

Delete Node in a BST 二叉查詢樹的查詢、插入和刪除 - Java實現

https://leetcode.com/problems/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

解題思路:
1. 根據BST的特性,找到需要刪除的node。
2. 如果node左右子樹都是空的,直接返回null。
3. 如果左子樹為空,那麼直接用右側子樹代替node。
4. 如果右側子樹為空,那麼直接用左子樹代替node。
5. 如果左右子樹都不是空,有些麻煩。首先要去node的右子樹中找到最小的那個節點,也就是右子樹中最左側的節點。
然後用它去替代node,並且把node的父節點的left置為空。
遞迴的時候,什麼時候直接return,什麼時候root.left = 呼叫遞迴,需要值得注意。
這個關係到,總是疑惑,刪除本node,那麼如何將父節點的left(right)置為null?需要好好理解。
詳細可以看
二叉查詢樹的查詢、插入和刪除 - Java實現
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null
) { return root; } if (root.val > key) { root.left = deleteNode(root.left, key); } else if (root.val < key) { root.right = deleteNode(root.right, key); } else { if (root.right == null) { return root.left; } else if (root.left == null) { return root.right; } else { TreeNode temp = root; root = findMin(root.right); root.right = deleteMin(temp.right); root.left = temp.left; } } return root; } private TreeNode findMin(TreeNode root) { if (root.left != null) { return findMin(root.left); } return root; } private TreeNode deleteMin(TreeNode root) { if (root.left != null) { root.left = deleteMin(root.left); } else { return root.right; } return root; } }