Delete Node in a BST

flyatcmu

已于 2022-05-04 07:22:43 修改

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分类专栏： Tree Divided & Conquer 文章标签： leetcode

于 2016-11-13 06:55:54 首次发布

本文链接：https://blog.youkuaiyun.com/u013325815/article/details/53146954

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本文介绍了一种在二叉搜索树中删除指定键的方法，包括查找待删除节点及删除节点的具体步骤。针对不同情况提供了详细的解决方案，并给出了两种有效的删除示例。

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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

思路：首先知道这个需要分情况讨论，三种情况：

1.左边空，return 右边

2.右边空，return 左边

3.左右两边都不空，找到最右边的最小值，然后把root改成最右边的最小值，然后delete那个最右边的最小node。

错误点：第三种情况，没有意识到，就是右边的都比自己大，选successor，作为自己的node，变动最小，然后delete successor即可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if(root == null) {
            return null;
        }
        if(root.val < key) {
            root.right = deleteNode(root.right, key);
        } else if(root.val > key) {
            root.left = deleteNode(root.left, key);
        } else {
            // root.val == key;
            if(root.left == null) {
                return root.right; 
            } else if(root.right == null) {
                return root.left;
            } else {
                TreeNode leftmostNode = findSuccessor(root.right);
                root.val = leftmostNode.val;
                root.right = deleteNode(root.right, leftmostNode.val);
            }
        }
        return root;
    }
    
    private TreeNode findSuccessor(TreeNode node) {
        while(node != null && node.left != null) {
            node = node.left;
        }
        return node;
    }
}