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chapter_divide_and_conquer/build_binary_tree_problem/index.html

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@@ -2276,7 +2292,7 @@
                   
 
   
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 <p class="admonition-title">Question</p>
 <p>给定一个二叉树的前序遍历 <code>preorder</code> 和中序遍历 <code>inorder</code> ，请从中构建二叉树，返回二叉树的根节点。</p>
 </div>
-<p><img alt="构建二叉树的示例数据" src="../build_binary_tree.assets/build_tree_example.png" /></p>
+<p><img alt="构建二叉树的示例数据" src="../build_binary_tree_problem.assets/build_tree_example.png" /></p>
 <p align="center"> Fig. 构建二叉树的示例数据 </p>
 
 <p>原问题定义为从 <code>preorder</code> 和 <code>inorder</code> 构建二叉树。我们首先从分治的角度分析这道题：</p>
@@ -2310,7 +2326,7 @@
 <li>查找根节点在 <code>inorder</code> 中的索引，基于该索引可将 <code>inorder</code> 划分为 <code>[ 9 | 3 ｜ 1 2 7 ]</code> ；</li>
 <li>根据 <code>inorder</code> 划分结果，可得左子树和右子树分别有 1 个和 3 个节点，从而可将 <code>preorder</code> 划分为 <code>[ 3 | 9 | 2 1 7 ]</code> ；</li>
 </ol>
-<p><img alt="在前序和中序遍历中划分子树" src="../build_binary_tree.assets/build_tree_preorder_inorder_division.png" /></p>
+<p><img alt="在前序和中序遍历中划分子树" src="../build_binary_tree_problem.assets/build_tree_preorder_inorder_division.png" /></p>
 <p align="center"> Fig. 在前序和中序遍历中划分子树 </p>
 
 <p>至此，<strong>我们已经推导出根节点、左子树、右子树在 <code>preorder</code> 和 <code>inorder</code> 中的索引区间</strong>。而为了描述这些索引区间，我们需要借助几个指针变量：</p>
@@ -2349,7 +2365,7 @@
 </table>
 </div>
 <p>请注意，右子树根节点索引中的 <span class="arithmatex">\((m-l)\)</span> 的含义是“左子树的节点数量”，建议配合下图理解。</p>
-<p><img alt="根节点和左右子树的索引区间表示" src="../build_binary_tree.assets/build_tree_division_pointers.png" /></p>
+<p><img alt="根节点和左右子树的索引区间表示" src="../build_binary_tree_problem.assets/build_tree_division_pointers.png" /></p>
 <p align="center"> Fig. 根节点和左右子树的索引区间表示 </p>
 
 <p>接下来就可以实现代码了。为了提升查询 <span class="arithmatex">\(m\)</span> 的效率，我们借助一个哈希表 <code>hmap</code> 来存储 <code>inorder</code> 列表元素到索引的映射。</p>
@@ -2376,7 +2392,7 @@
 <a id="__codelineno-2-6" name="__codelineno-2-6" href="#__codelineno-2-6"></a>    <span class="n">l</span><span class="p">:</span> <span class="nb">int</span><span class="p">,</span>
 <a id="__codelineno-2-7" name="__codelineno-2-7" href="#__codelineno-2-7"></a>    <span class="n">r</span><span class="p">:</span> <span class="nb">int</span><span class="p">,</span>
 <a id="__codelineno-2-8" name="__codelineno-2-8" href="#__codelineno-2-8"></a><span class="p">)</span> <span class="o">-&gt;</span> <span class="n">TreeNode</span> <span class="o">|</span> <span class="kc">None</span><span class="p">:</span>
-<a id="__codelineno-2-9" name="__codelineno-2-9" href="#__codelineno-2-9"></a><span class="w">    </span><span class="sd">&quot;&quot;&quot;构建二叉树 DFS&quot;&quot;&quot;</span>
+<a id="__codelineno-2-9" name="__codelineno-2-9" href="#__codelineno-2-9"></a><span class="w">    </span><span class="sd">&quot;&quot;&quot;构建二叉树：分治&quot;&quot;&quot;</span>
 <a id="__codelineno-2-10" name="__codelineno-2-10" href="#__codelineno-2-10"></a>    <span class="c1"># 子树区间为空时终止</span>
 <a id="__codelineno-2-11" name="__codelineno-2-11" href="#__codelineno-2-11"></a>    <span class="k">if</span> <span class="n">r</span> <span class="o">-</span> <span class="n">l</span> <span class="o">&lt;</span> <span class="mi">0</span><span class="p">:</span>
 <a id="__codelineno-2-12" name="__codelineno-2-12" href="#__codelineno-2-12"></a>        <span class="k">return</span> <span class="kc">None</span>
@@ -2384,9 +2400,9 @@
 <a id="__codelineno-2-14" name="__codelineno-2-14" href="#__codelineno-2-14"></a>    <span class="n">root</span> <span class="o">=</span> <span class="n">TreeNode</span><span class="p">(</span><span class="n">preorder</span><span class="p">[</span><span class="n">i</span><span class="p">])</span>
 <a id="__codelineno-2-15" name="__codelineno-2-15" href="#__codelineno-2-15"></a>    <span class="c1"># 查询 m ，从而划分左右子树</span>
 <a id="__codelineno-2-16" name="__codelineno-2-16" href="#__codelineno-2-16"></a>    <span class="n">m</span> <span class="o">=</span> <span class="n">hmap</span><span class="p">[</span><span class="n">preorder</span><span class="p">[</span><span class="n">i</span><span class="p">]]</span>
-<a id="__codelineno-2-17" name="__codelineno-2-17" href="#__codelineno-2-17"></a>    <span class="c1"># 递归构建左子树</span>
+<a id="__codelineno-2-17" name="__codelineno-2-17" href="#__codelineno-2-17"></a>    <span class="c1"># 子问题：构建左子树</span>
 <a id="__codelineno-2-18" name="__codelineno-2-18" href="#__codelineno-2-18"></a>    <span class="n">root</span><span class="o">.</span><span class="n">left</span> <span class="o">=</span> <span class="n">dfs</span><span class="p">(</span><span class="n">preorder</span><span class="p">,</span> <span class="n">inorder</span><span class="p">,</span> <span class="n">hmap</span><span class="p">,</span> <span class="n">i</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">l</span><span class="p">,</span> <span class="n">m</span> <span class="o">-</span> <span class="mi">1</span><span class="p">)</span>
-<a id="__codelineno-2-19" name="__codelineno-2-19" href="#__codelineno-2-19"></a>    <span class="c1"># 递归构建右子树</span>
+<a id="__codelineno-2-19" name="__codelineno-2-19" href="#__codelineno-2-19"></a>    <span class="c1"># 子问题：构建右子树</span>
 <a id="__codelineno-2-20" name="__codelineno-2-20" href="#__codelineno-2-20"></a>    <span class="n">root</span><span class="o">.</span><span class="n">right</span> <span class="o">=</span> <span class="n">dfs</span><span class="p">(</span><span class="n">preorder</span><span class="p">,</span> <span class="n">inorder</span><span class="p">,</span> <span class="n">hmap</span><span class="p">,</span> <span class="n">i</span> <span class="o">+</span> <span class="mi">1</span> <span class="o">+</span> <span class="n">m</span> <span class="o">-</span> <span class="n">l</span><span class="p">,</span> <span class="n">m</span> <span class="o">+</span> <span class="mi">1</span><span class="p">,</span> <span class="n">r</span><span class="p">)</span>
 <a id="__codelineno-2-21" name="__codelineno-2-21" href="#__codelineno-2-21"></a>    <span class="c1"># 返回根节点</span>
 <a id="__codelineno-2-22" name="__codelineno-2-22" href="#__codelineno-2-22"></a>    <span class="k">return</span> <span class="n">root</span>
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 <div class="tabbed-content">
 <div class="tabbed-block">
-<p><img alt="built_tree_step1" src="../build_binary_tree.assets/built_tree_step1.png" /></p>
+<p><img alt="构建二叉树的递归过程" src="../build_binary_tree_problem.assets/built_tree_step1.png" /></p>
 </div>
 <div class="tabbed-block">
-<p><img alt="built_tree_step2" src="../build_binary_tree.assets/built_tree_step2.png" /></p>
+<p><img alt="built_tree_step2" src="../build_binary_tree_problem.assets/built_tree_step2.png" /></p>
 </div>
 <div class="tabbed-block">
-<p><img alt="built_tree_step3" src="../build_binary_tree.assets/built_tree_step3.png" /></p>
+<p><img alt="built_tree_step3" src="../build_binary_tree_problem.assets/built_tree_step3.png" /></p>
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-<p><img alt="built_tree_step5" src="../build_binary_tree.assets/built_tree_step5.png" /></p>
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-<p><img alt="built_tree_step7" src="../build_binary_tree.assets/built_tree_step7.png" /></p>
+<p><img alt="built_tree_step7" src="../build_binary_tree_problem.assets/built_tree_step7.png" /></p>
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-<p><img alt="built_tree_step8" src="../build_binary_tree.assets/built_tree_step8.png" /></p>
+<p><img alt="built_tree_step8" src="../build_binary_tree_problem.assets/built_tree_step8.png" /></p>
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-<p><img alt="built_tree_step9" src="../build_binary_tree.assets/built_tree_step9.png" /></p>
+<p><img alt="built_tree_step9" src="../build_binary_tree_problem.assets/built_tree_step9.png" /></p>
 </div>
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-<p><img alt="built_tree_step10" src="../build_binary_tree.assets/built_tree_step10.png" /></p>
+<p><img alt="built_tree_step10" src="../build_binary_tree_problem.assets/built_tree_step10.png" /></p>
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@@ -2579,13 +2595,13 @@
         
         
           
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chapter_divide_and_conquer/divide_and_conquer/index.html

@@ -18,7 +18,7 @@
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-      <a href="../build_binary_tree/" class="md-nav__link">
+      <a href="../build_binary_tree_problem/" class="md-nav__link">
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@@ -1891,6 +1893,20 @@
 
             
           
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+      <a href="../hanota_problem/" class="md-nav__link">
+        12.3. &nbsp; 汉诺塔问题（New）
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@@ -2384,7 +2400,7 @@
   
 
 
-<h1 id="121">12.1. &nbsp; 分治<a class="headerlink" href="#121" title="Permanent link">&para;</a></h1>
+<h1 id="121">12.1. &nbsp; 分治算法<a class="headerlink" href="#121" title="Permanent link">&para;</a></h1>
 <p>「分治 Divide and Conquer」，全称分而治之，是一种非常重要的算法策略。分治通常基于递归实现，包括“分”和“治”两部分，主要步骤如下：</p>
 <ol>
 <li><strong>分（划分阶段）</strong>：递归地将原问题分解为两个或多个子问题，直至到达最小子问题时终止；</li>
@@ -2552,7 +2568,7 @@ n(n - 4) &amp; &gt; 0
         
         
           
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@@ -1823,6 +1825,20 @@
 
             
           
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