22天【程式碼隨想錄演算法訓練營34期】第六章 二叉樹part08 （● 235. 二叉搜尋樹的最近公共祖先 ● 701.二叉搜尋樹中的插入操作 ● 450.刪除二叉搜尋樹中的節點）

MiraMira發表於2024-04-10
演算法二叉樹

235. 二叉搜尋樹的最近公共祖先
因為是搜尋二叉樹，所以只要值在q和p之間，那麼就是lowest common ancestor

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

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.right, p, q)
        if root.val > p.val and root.val > q.val:
            return self.lowestCommonAncestor(root.left, p, q)
        return root

701.二叉搜尋樹中的插入操作

# 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 insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
        if root is None:
            return TreeNode(val)
        elif root.val > val:
            root.left = self.insertIntoBST(root.left, val)
        elif root.val < val:
            root.right = self.insertIntoBST(root.right, val)
        return root

450.刪除二叉搜尋樹中的節點
怎麼刪除一個根結點，將它的值和right subtree的最左葉子的值交換，然後刪除最左葉子

# 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:
            if root.right is None:
                return root.left
            cur = root.right
            while cur.left:
                cur = cur.left
            cur.val, root.val = root.val, cur.val
        root.left = self.deleteNode(root.left, key)
        root.right = self.deleteNode(root.right, key)
        return root

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