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450. 删除二叉搜索树中的节点

English Version

题目描述

给定一个二叉搜索树的根节点 root 和一个值 key，删除二叉搜索树中的 key 对应的节点，并保证二叉搜索树的性质不变。返回二叉搜索树（有可能被更新）的根节点的引用。

一般来说，删除节点可分为两个步骤：

1. 首先找到需要删除的节点；
2. 如果找到了，删除它。

说明： 要求算法时间复杂度为 O(h)，h 为树的高度。

示例:

root = [5,3,6,2,4,null,7]
key = 3

    5
   / \
  3   6
 / \   \
2   4   7

给定需要删除的节点值是 3，所以我们首先找到 3 这个节点，然后删除它。

一个正确的答案是 [5,4,6,2,null,null,7], 如下图所示。

    5
   / \
  4   6
 /     \
2       7

另一个正确答案是 [5,2,6,null,4,null,7]。

    5
   / \
  2   6
   \   \
    4   7

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
        if root is None:
            return None
        if root.val > key:
            root.left = self.deleteNode(root.left, key)
            return root
        if root.val < key:
            root.right = self.deleteNode(root.right, key)
            return root
        if root.left is None:
            return root.right
        if root.right is None:
            return root.left
        node = root.right
        while node.left:
            node = node.left
        node.left = root.left
        root = root.right
        return root

Java

/**
 * 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.left = deleteNode(root.left, key);
            return root;
        }
        if (root.val < key) {
            root.right = deleteNode(root.right, key);
            return root;
        }
        if (root.left == null) {
            return root.right;
        }
        if (root.right == null) {
            return root.left;
        }
        TreeNode node = root.right;
        while (node.left != null) {
            node = node.left;
        }
        node.left = root.left;
        root = root.right;
        return root;
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* deleteNode(TreeNode* root, int key) {
        if (!root) return root;
        if (root->val > key)
        {
            root->left = deleteNode(root->left, key);
            return root;
        }
        if (root->val < key)
        {
            root->right = deleteNode(root->right, key);
            return root;
        }
        if (!root->left) return root->right;
        if (!root->right) return root->left;
        TreeNode* node = root->right;
        while (node->left) node = node->left;
        node->left = root->left;
        root = root->right;
        return root;
    }
};

Go

/**
 * 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 nil
	}
	if root.Val > key {
		root.Left = deleteNode(root.Left, key)
		return root
	}
	if root.Val < key {
		root.Right = deleteNode(root.Right, key)
		return root
	}
	if root.Left == nil {
		return root.Right
	}
	if root.Right == nil {
		return root.Left
	}
	node := root.Right
	for node.Left != nil {
		node = node.Left
	}
	node.Left = root.Left
	root = root.Right
	return root
}

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