代码随想训练营第二十二天（Python）｜ 235. 二叉搜索树的最近公共祖先、701.二叉搜索树中的插入操作、450.删除二叉搜索树中的节点

235. 二叉搜索树的最近公共祖先
关键点：最近公共祖先的判断，二叉树的特性
1、做二叉树的模式

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if root is None:
            return None
        if root == p or root == q:
            return root

        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)

        if left and right:
            return root

        if not left:
            return right

        return left

2、二叉搜索树优化,只需要判断一边就可以返回

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if root is None:
            return None
        if root.val > p.val and root.val > q.val:
            return self.lowestCommonAncestor(root.left, p, q)
        elif root.val < p.val and root.val < q.val:
            return self.lowestCommonAncestor(root.right, p, q)
        else:
            return root

3、迭代法

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        while root:
            if root.val > p.val and root.val > q.val:
                root = root.left
            elif root.val < p.val and root.val < q.val:
                root = root.right
            else:
                return root
        return None

701.二叉搜索树中的插入操作
赋值需要知道父节点，需要一个指针
1、递归法

class Solution:
    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        if root is None:
            return TreeNode(val)

        if root.val > val:
            root.left = self.insertIntoBST(root.left, val)
        if root.val < val:
            root.right = self.insertIntoBST(root.right, val)
        return root

2、迭代法+指针

class Solution:
    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        if root is None:
            return TreeNode(val)
        cur = root
        pre = None
        while cur:
            pre = cur
            if cur.val > val:
                cur = cur.left
            else:
                cur = cur.right
        if pre.val > val:
            pre.left = TreeNode(val)
        else:
            pre.right = TreeNode(val)

        return root

450.删除二叉搜索树中的节点
1、递归法

class Solution:
    def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
        if root is None:
            return None
        if root.val == key:
            if not root.right:
                return root.left
            if not root.left:
                return root.right
            cur = root.right
            while cur.left:
                cur = cur.left
            cur.left = root.left
            return root.right
        if root.val > key:
            root.left = self.deleteNode(root.left, key)
        else:
            root.right = self.deleteNode(root.right, key)
        return root

2、公共二叉树删除法。选出右节点最左边树的值与删除节点交换

class Solution:
    def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]:
        if root is None:
            return None
        if root.val == key:
            if not root.right:
                return root.left  # root.left 为 None 也可以返回
            cur = root.right
            while cur.left:
                cur = cur.left
            root.val, cur.val = cur.val, root.val  # 交换右子树最左节点和根节点的值。

            # 后序会遍历到 root.val == key 的逻辑，root.left 与 root.right 都为 None, 可以返回 None。相当与删除了节点
            
        root.left = self.deleteNode(root.left, key)
        root.right = self.deleteNode(root.right, key)
        return root