Leetcode 450.删除二叉搜索树中的节点

Posted on |

450.删除二叉搜索树中的节点

题目

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：

首先找到需要删除的节点；
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:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
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

方法

方法1：递归

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
# 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 root is None:
            return

        if root.val==key:
            if not root.left and not root.right:
                return None
            elif not root.right:
                return root.left
            elif not root.left:
                return root.right
            else:
                root.val=self.successor(root.right)
                root.right=self.deleteNode(root.right,root.val)
        elif root.val<key:
            root.right=self.deleteNode(root.right,key)
        else:
            root.left=self.deleteNode(root.left,key)

        return root

    def successor(self,root):
        while root.left:
            root=root.left
        return root.val
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def deleteNode(self, root, key):
        """
        :type root: TreeNode
        :type key: int
        :rtype: TreeNode
        """
        if not root:
            return

        # we always want to delete the node when it is the root of a subtree,
        # so we handle left or right according to the val.
        # if the node does not exist, we will hit the very first if statement and return None.
        if key > root.val:
            root.right = self.deleteNode(root.right, key)

        elif key < root.val:
            root.left = self.deleteNode(root.left, key)

        # now the key is the root of a subtree
        else:
            # if the subtree does not have a left child, we just return its right child
            # to its father, and they will be connected on the higher level recursion.
            if not root.left:
                return root.right

            # if it has a left child, we want to find the max val on the left subtree to
            # replace the node we want to delete.
            else:
                # try to find the max value on the left subtree
                tmp = root.left
                while tmp.right:
                    tmp = tmp.right

                # replace
                root.val = tmp.val

                # since we have replaced the node we want to delete with the tmp, now we don't
                # want to keep the tmp on this tree, so we just use our function to delete it.
                # pass the val of tmp to the left subtree and repeat the whole approach.
                root.left = self.deleteNode(root.left, tmp.val)

        return root