From dac0b4a12e43670eb54eaa6da5fa3f07cb707424 Mon Sep 17 00:00:00 2001 From: GitHubQAQ <31883473+GitHubQAQ@users.noreply.github.com> Date: Fri, 29 Apr 2022 21:03:27 +0800 Subject: [PATCH] =?UTF-8?q?Update=200106.=E4=BB=8E=E4=B8=AD=E5=BA=8F?= =?UTF-8?q?=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86=E5=BA=8F=E5=88=97?= =?UTF-8?q?=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 优化代码高亮 --- .../0106.从中序与后序遍历序列构造二叉树.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index 496de431..4396bc76 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -103,7 +103,7 @@ TreeNode* traversal (vector& inorder, vector& postorder) { 中序数组相对比较好切，找到切割点（后序数组的最后一个元素）在中序数组的位置，然后切割，如下代码中我坚持左闭右开的原则： -```C++ +```CPP // 找到中序遍历的切割点 int delimiterIndex; for (delimiterIndex = 0; delimiterIndex < inorder.size(); delimiterIndex++) { @@ -130,7 +130,7 @@ vector rightInorder(inorder.begin() + delimiterIndex + 1, inorder.end() ); 代码如下： -``` +```CPP // postorder 舍弃末尾元素，因为这个元素就是中间节点，已经用过了 postorder.resize(postorder.size() - 1); @@ -144,7 +144,7 @@ vector rightPostorder(postorder.begin() + leftInorder.size(), postorder.end 接下来可以递归了，代码如下： -``` +```CPP root->left = traversal(leftInorder, leftPostorder); root->right = traversal(rightInorder, rightPostorder); ```