From 731d1ca8aa61b1b8969ead0c5cf8b7d8b21b55e6 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 21:38:27 -0500 Subject: [PATCH] =?UTF-8?q?Update=200450.=E5=88=A0=E9=99=A4=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E8=8A=82?= =?UTF-8?q?=E7=82=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0450.删除二叉搜索树中的节点.md | 132 +++++++++--------- 1 file changed, 67 insertions(+), 65 deletions(-) diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md index 2602b528..d4dca1d4 100644 --- a/problems/0450.删除二叉搜索树中的节点.md +++ b/problems/0450.删除二叉搜索树中的节点.md @@ -324,88 +324,90 @@ class Solution { ``` ## Python - +递归法(版本一) ```python class Solution: - def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]: - if not root : return None # 节点为空,返回 - if root.val < key : - root.right = self.deleteNode(root.right, key) - elif root.val > key : + def deleteNode(self, root, key): + if root is None: + return root + if root.val == key: + if root.left is None and root.right is None: + return None + elif root.left is None: + return root.right + elif root.right is None: + return root.left + else: + cur = root.right + while cur.left is not None: + cur = cur.left + cur.left = root.left + return root.right + if root.val > key: root.left = self.deleteNode(root.left, key) - else: - # 当前节点的左子树为空,返回当前的右子树 - if not root.left : return root.right - # 当前节点的右子树为空,返回当前的左子树 - if not root.right: return root.left - # 左右子树都不为空,找到右孩子的最左节点 记为p - node = root.right - while node.left : - node = node.left - # 将当前节点的左子树挂在p的左孩子上 - node.left = root.left - # 当前节点的右子树替换掉当前节点,完成当前节点的删除 - root = root.right + if root.val < key: + root.right = self.deleteNode(root.right, key) return root ``` -**普通二叉树的删除方式** +递归法(版本二) ```python class Solution: - def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]: - if not root: return root - if root.val == key: - if not root.right: # 这里第二次操作目标值:最终删除的作用 + def deleteNode(self, root, key): + if root is None: # 如果根节点为空,直接返回 + return root + if root.val == key: # 找到要删除的节点 + if root.right is None: # 如果右子树为空,直接返回左子树作为新的根节点 return root.left - tmp = root.right - while tmp.left: - tmp = tmp.left - root.val, tmp.val = tmp.val, root.val # 这里第一次操作目标值:交换目标值其右子树最左面节点。 - - root.left = self.deleteNode(root.left, key) - root.right = self.deleteNode(root.right, key) + cur = root.right + while cur.left: # 找到右子树中的最左节点 + cur = cur.left + root.val, cur.val = cur.val, root.val # 将要删除的节点值与最左节点值交换 + root.left = self.deleteNode(root.left, key) # 在左子树中递归删除目标节点 + root.right = self.deleteNode(root.right, key) # 在右子树中递归删除目标节点 return root + ``` **迭代法** ```python class Solution: - def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]: - # 找到节点后分两步,1. 把节点的左子树和右子树连起来,2. 把右子树跟父节点连起来 - # root is None - if not root: return root - p = root - last = None - while p: - if p.val==key: - # 1. connect left to right - # right is not None -> left is None | left is not None - if p.right: - if p.left: - node = p.right - while node.left: - node = node.left - node.left = p.left - right = p.right - else: - # right is None -> right=left - right = p.left - # 2. connect right to last - if last==None: - root = right - elif last.val>key: - last.left = right - else: - last.right = right - # 3. return + def deleteOneNode(self, target: TreeNode) -> TreeNode: + """ + 将目标节点(删除节点)的左子树放到目标节点的右子树的最左面节点的左孩子位置上 + 并返回目标节点右孩子为新的根节点 + 是动画里模拟的过程 + """ + if target is None: + return target + if target.right is None: + return target.left + cur = target.right + while cur.left: + cur = cur.left + cur.left = target.left + return target.right + + def deleteNode(self, root: TreeNode, key: int) -> TreeNode: + if root is None: + return root + cur = root + pre = None # 记录cur的父节点,用来删除cur + while cur: + if cur.val == key: break + pre = cur + if cur.val > key: + cur = cur.left else: - # Update last and continue - last = p - if p.val>key: - p = p.left - else: - p = p.right + cur = cur.right + if pre is None: # 如果搜索树只有头结点 + return self.deleteOneNode(cur) + # pre 要知道是删左孩子还是右孩子 + if pre.left and pre.left.val == key: + pre.left = self.deleteOneNode(cur) + if pre.right and pre.right.val == key: + pre.right = self.deleteOneNode(cur) return root ```