题目地址:https://leetcode.com/problems/delete-node-in-a-bst/description/

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

题目大意

删除二叉树中指定一节点,并调整二叉树,使得结果的二叉树仍然满足BST的条件。

解题方法

迭代

这个题的解法并不是固定的,删除之后的二叉树也不止一种。比如可以有下面两种主要的方法:

1、 被删除节点没有左子树:返回其右子树;
2、 被删除节点节点没有右子树:返回其左子树;
3、 被删除节点既有左子树,又有右子树:1)查找到其右子树的最小值的节点,替换掉被删除的节点,并删除找到的最小节点2)查找到其左子树的最大值的节点,替换掉被删除的节点,并删除找到的最大节点;

下面的做法是查找右子树的最小值节点的方法,最小节点就是右子树中的最靠左边的节点。代码使用的递归,最核心的是找到该节点之后的操作,特别是把值进行交换一步很重要,因为我们并没有删除了该最小值节点,所以把最小值的节点赋值成要查找的节点,然后在之后的操作中将会把它删除。

Python代码如下:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def deleteNode(self, root, key):
        """
        :type root: TreeNode
        :type key: int
        :rtype: TreeNode
        """
        if not root: return None
        if root.val == key:
            if not root.right:
                left = root.left
                return left
            else:
                right = root.right
                while right.left:
                    right = right.left
                root.val, right.val = right.val, root.val
        root.left = self.deleteNode(root.left, key)
        root.right = self.deleteNode(root.right, key)
        return root

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删除查找到左子树中最大值的方法,左子树的最大值是左子树的最靠右边的节点,使用迭代找到该值,然后和root节点的值进行交换。递归左右子树,这个被交换的值会在后面被删除掉。

C++代码如下:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return nullptr;
        if (root->val == key) {
            if (!root->left) {
                return root->right;
            } else {
                TreeNode* left = root->left;
                while (left->right) {
                    left = left->right;
                }
                swap(left->val, root->val);
            }
        }
        root->left = deleteNode(root->left, key);
        root->right = deleteNode(root->right, key);
        return root;
    }
};

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