0450. Delete Node in a BST (M)

Delete Node in a BST (M)

题目

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

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题意

删除二叉搜索树中的一个结点。

思路

使用递归实现：

1. 如果 key<root.val，向左递归;
2. 如果 key>root.val，向右递归；
3. 如果 key==root.val，分为3中情况：
   1. root没有左子树，那么直接返回root的右子树；
   2. root没有右子树，那么直接返回root的左子树；
   3. root既有左子树又有右子树，那么在root的右子树中找到最小的值，即右子树中最左侧的结点，将它的值赋给root，再删掉这个结点(即上述没有左子树的情况)。也可以找root左子树中最大(最右侧)的结点。

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代码实现

Java

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) {
            root = root.right;
        } else if (root.right == null) {
            root = root.left;
        } else {
            TreeNode cur = root.right;
            TreeNode pre = root;
            while (cur.left != null) {
                cur = cur.left;
                pre = pre == root ? root.right : pre.left;
            }
            root.val = cur.val;
            if (pre == root) {
                pre.right = cur.right;
            } else {
                pre.left = cur.right;
            }
        }

        return root;
    }
}