From a518db03dd3265f53d98d2277f5da12323363958 Mon Sep 17 00:00:00 2001 From: sss1h <49825610+sss1h@users.noreply.github.com> Date: Tue, 27 Feb 2024 20:26:06 +0800 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 添加代码高亮 --- problems/0450.删除二叉搜索树中的节点.md | 6 +++--- 1 file changed, 3 insertions(+), 3 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 8922a14e..60dae7b9 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 @@ -42,7 +42,7 @@ 代码如下: -``` +```cpp TreeNode* deleteNode(TreeNode* root, int key) ``` @@ -50,7 +50,7 @@ TreeNode* deleteNode(TreeNode* root, int key) 遇到空返回,其实这也说明没找到删除的节点,遍历到空节点直接返回了 -``` +```cpp if (root == nullptr) return root; ``` @@ -106,7 +106,7 @@ if (root->val == key) { 这里相当于把新的节点返回给上一层,上一层就要用 root->left 或者 root->right接住,代码如下: -``` +```cpp if (root->val > key) root->left = deleteNode(root->left, key); if (root->val < key) root->right = deleteNode(root->right, key); return root;