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chapter_tree/avl_tree/index.html

@@ -3619,11 +3619,11 @@
 <p>在二叉搜索树章节中，我们提到了在多次插入和删除操作后，二叉搜索树可能退化为链表。这种情况下，所有操作的时间复杂度将从 <span class="arithmatex">\(O(\log n)\)</span> 恶化为 <span class="arithmatex">\(O(n)\)</span>。</p>
 <p>如下图所示，经过两次删除节点操作，这个二叉搜索树便会退化为链表。</p>
 <p><img alt="AVL 树在删除节点后发生退化" src="../avl_tree.assets/avltree_degradation_from_removing_node.png" /></p>
-<p align="center"> Fig. AVL 树在删除节点后发生退化 </p>
+<p align="center"> 图：AVL 树在删除节点后发生退化 </p>
 
 <p>再例如，在以下完美二叉树中插入两个节点后，树将严重向左倾斜，查找操作的时间复杂度也随之恶化。</p>
 <p><img alt="AVL 树在插入节点后发生退化" src="../avl_tree.assets/avltree_degradation_from_inserting_node.png" /></p>
-<p align="center"> Fig. AVL 树在插入节点后发生退化 </p>
+<p align="center"> 图：AVL 树在插入节点后发生退化 </p>
 
 <p>G. M. Adelson-Velsky 和 E. M. Landis 在其 1962 年发表的论文 "An algorithm for the organization of information" 中提出了「AVL 树」。论文中详细描述了一系列操作，确保在持续添加和删除节点后，AVL 树不会退化，从而使得各种操作的时间复杂度保持在 <span class="arithmatex">\(O(\log n)\)</span> 级别。换句话说，在需要频繁进行增删查改操作的场景中，AVL 树能始终保持高效的数据操作性能，具有很好的应用价值。</p>
 <h2 id="751-avl">7.5.1. &nbsp; AVL 树常见术语<a class="headerlink" href="#751-avl" title="Permanent link">&para;</a></h2>
@@ -3945,14 +3945,15 @@
 <div class="tabbed-block">
 <div class="highlight"><span class="filename">avl_tree.dart</span><pre><span></span><code><a id="__codelineno-22-1" name="__codelineno-22-1" href="#__codelineno-22-1"></a><span class="cm">/* 获取节点高度 */</span>
 <a id="__codelineno-22-2" name="__codelineno-22-2" href="#__codelineno-22-2"></a><span class="kt">int</span><span class="w"> </span><span class="n">height</span><span class="p">(</span><span class="n">TreeNode</span><span class="o">?</span><span class="w"> </span><span class="n">node</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-22-3" name="__codelineno-22-3" href="#__codelineno-22-3"></a><span class="w">  </span><span class="k">return</span><span class="w"> </span><span class="n">node</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="kc">null</span><span class="w"> </span><span class="o">?</span><span class="w"> </span><span class="o">-</span><span class="m">1</span><span class="w"> </span><span class="o">:</span><span class="w"> </span><span class="n">node</span><span class="p">.</span><span class="n">height</span><span class="p">;</span>
-<a id="__codelineno-22-4" name="__codelineno-22-4" href="#__codelineno-22-4"></a><span class="p">}</span>
-<a id="__codelineno-22-5" name="__codelineno-22-5" href="#__codelineno-22-5"></a>
-<a id="__codelineno-22-6" name="__codelineno-22-6" href="#__codelineno-22-6"></a><span class="cm">/* 更新节点高度 */</span>
-<a id="__codelineno-22-7" name="__codelineno-22-7" href="#__codelineno-22-7"></a><span class="kt">void</span><span class="w"> </span><span class="n">updateHeight</span><span class="p">(</span><span class="n">TreeNode</span><span class="o">?</span><span class="w"> </span><span class="n">node</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
-<a id="__codelineno-22-8" name="__codelineno-22-8" href="#__codelineno-22-8"></a><span class="w">  </span><span class="c1">// 节点高度等于最高子树高度 + 1</span>
-<a id="__codelineno-22-9" name="__codelineno-22-9" href="#__codelineno-22-9"></a><span class="w">  </span><span class="n">node</span><span class="o">!</span><span class="p">.</span><span class="n">height</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">max</span><span class="p">(</span><span class="n">height</span><span class="p">(</span><span class="n">node</span><span class="p">.</span><span class="n">left</span><span class="p">),</span><span class="w"> </span><span class="n">height</span><span class="p">(</span><span class="n">node</span><span class="p">.</span><span class="n">right</span><span class="p">))</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="m">1</span><span class="p">;</span>
-<a id="__codelineno-22-10" name="__codelineno-22-10" href="#__codelineno-22-10"></a><span class="p">}</span>
+<a id="__codelineno-22-3" name="__codelineno-22-3" href="#__codelineno-22-3"></a><span class="w">  </span><span class="c1">// 空节点高度为 -1 ，叶节点高度为 0</span>
+<a id="__codelineno-22-4" name="__codelineno-22-4" href="#__codelineno-22-4"></a><span class="w">  </span><span class="k">return</span><span class="w"> </span><span class="n">node</span><span class="w"> </span><span class="o">==</span><span class="w"> </span><span class="kc">null</span><span class="w"> </span><span class="o">?</span><span class="w"> </span><span class="o">-</span><span class="m">1</span><span class="w"> </span><span class="o">:</span><span class="w"> </span><span class="n">node</span><span class="p">.</span><span class="n">height</span><span class="p">;</span>
+<a id="__codelineno-22-5" name="__codelineno-22-5" href="#__codelineno-22-5"></a><span class="p">}</span>
+<a id="__codelineno-22-6" name="__codelineno-22-6" href="#__codelineno-22-6"></a>
+<a id="__codelineno-22-7" name="__codelineno-22-7" href="#__codelineno-22-7"></a><span class="cm">/* 更新节点高度 */</span>
+<a id="__codelineno-22-8" name="__codelineno-22-8" href="#__codelineno-22-8"></a><span class="kt">void</span><span class="w"> </span><span class="n">updateHeight</span><span class="p">(</span><span class="n">TreeNode</span><span class="o">?</span><span class="w"> </span><span class="n">node</span><span class="p">)</span><span class="w"> </span><span class="p">{</span>
+<a id="__codelineno-22-9" name="__codelineno-22-9" href="#__codelineno-22-9"></a><span class="w">  </span><span class="c1">// 节点高度等于最高子树高度 + 1</span>
+<a id="__codelineno-22-10" name="__codelineno-22-10" href="#__codelineno-22-10"></a><span class="w">  </span><span class="n">node</span><span class="o">!</span><span class="p">.</span><span class="n">height</span><span class="w"> </span><span class="o">=</span><span class="w"> </span><span class="n">max</span><span class="p">(</span><span class="n">height</span><span class="p">(</span><span class="n">node</span><span class="p">.</span><span class="n">left</span><span class="p">),</span><span class="w"> </span><span class="n">height</span><span class="p">(</span><span class="n">node</span><span class="p">.</span><span class="n">right</span><span class="p">))</span><span class="w"> </span><span class="o">+</span><span class="w"> </span><span class="m">1</span><span class="p">;</span>
+<a id="__codelineno-22-11" name="__codelineno-22-11" href="#__codelineno-22-11"></a><span class="p">}</span>
 </code></pre></div>
 </div>
 <div class="tabbed-block">
@@ -4139,9 +4140,11 @@
 </div>
 </div>
 </div>
+<p align="center"> 图：右旋操作步骤 </p>
+
 <p>此外，如果节点 <code>child</code> 本身有右子节点（记为 <code>grandChild</code> ），则需要在「右旋」中添加一步：将 <code>grandChild</code> 作为 <code>node</code> 的左子节点。</p>
 <p><img alt="有 grandChild 的右旋操作" src="../avl_tree.assets/avltree_right_rotate_with_grandchild.png" /></p>
-<p align="center"> Fig. 有 grandChild 的右旋操作 </p>
+<p align="center"> 图：有 grandChild 的右旋操作 </p>
 
 <p>“向右旋转”是一种形象化的说法，实际上需要通过修改节点指针来实现，代码如下所示。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="5:12"><input checked="checked" id="__tabbed_5_1" name="__tabbed_5" type="radio" /><input id="__tabbed_5_2" name="__tabbed_5" type="radio" /><input id="__tabbed_5_3" name="__tabbed_5" type="radio" /><input id="__tabbed_5_4" name="__tabbed_5" type="radio" /><input id="__tabbed_5_5" name="__tabbed_5" type="radio" /><input id="__tabbed_5_6" name="__tabbed_5" type="radio" /><input id="__tabbed_5_7" name="__tabbed_5" type="radio" /><input id="__tabbed_5_8" name="__tabbed_5" type="radio" /><input id="__tabbed_5_9" name="__tabbed_5" type="radio" /><input id="__tabbed_5_10" name="__tabbed_5" type="radio" /><input id="__tabbed_5_11" name="__tabbed_5" type="radio" /><input id="__tabbed_5_12" name="__tabbed_5" type="radio" /><div class="tabbed-labels"><label for="__tabbed_5_1">Java</label><label for="__tabbed_5_2">C++</label><label for="__tabbed_5_3">Python</label><label for="__tabbed_5_4">Go</label><label for="__tabbed_5_5">JS</label><label for="__tabbed_5_6">TS</label><label for="__tabbed_5_7">C</label><label for="__tabbed_5_8">C#</label><label for="__tabbed_5_9">Swift</label><label for="__tabbed_5_10">Zig</label><label for="__tabbed_5_11">Dart</label><label for="__tabbed_5_12">Rust</label></div>
@@ -4348,11 +4351,11 @@
 <h3 id="_4">左旋<a class="headerlink" href="#_4" title="Permanent link">&para;</a></h3>
 <p>相应的，如果考虑上述失衡二叉树的“镜像”，则需要执行「左旋」操作。</p>
 <p><img alt="左旋操作" src="../avl_tree.assets/avltree_left_rotate.png" /></p>
-<p align="center"> Fig. 左旋操作 </p>
+<p align="center"> 图：左旋操作 </p>
 
 <p>同理，若节点 <code>child</code> 本身有左子节点（记为 <code>grandChild</code> ），则需要在「左旋」中添加一步：将 <code>grandChild</code> 作为 <code>node</code> 的右子节点。</p>
 <p><img alt="有 grandChild 的左旋操作" src="../avl_tree.assets/avltree_left_rotate_with_grandchild.png" /></p>
-<p align="center"> Fig. 有 grandChild 的左旋操作 </p>
+<p align="center"> 图：有 grandChild 的左旋操作 </p>
 
 <p>可以观察到，<strong>右旋和左旋操作在逻辑上是镜像对称的，它们分别解决的两种失衡情况也是对称的</strong>。基于对称性，我们可以轻松地从右旋的代码推导出左旋的代码。具体地，只需将「右旋」代码中的把所有的 <code>left</code> 替换为 <code>right</code> ，将所有的 <code>right</code> 替换为 <code>left</code> ，即可得到「左旋」代码。</p>
 <div class="tabbed-set tabbed-alternate" data-tabs="6:12"><input checked="checked" id="__tabbed_6_1" name="__tabbed_6" type="radio" /><input id="__tabbed_6_2" name="__tabbed_6" type="radio" /><input id="__tabbed_6_3" name="__tabbed_6" type="radio" /><input id="__tabbed_6_4" name="__tabbed_6" type="radio" /><input id="__tabbed_6_5" name="__tabbed_6" type="radio" /><input id="__tabbed_6_6" name="__tabbed_6" type="radio" /><input id="__tabbed_6_7" name="__tabbed_6" type="radio" /><input id="__tabbed_6_8" name="__tabbed_6" type="radio" /><input id="__tabbed_6_9" name="__tabbed_6" type="radio" /><input id="__tabbed_6_10" name="__tabbed_6" type="radio" /><input id="__tabbed_6_11" name="__tabbed_6" type="radio" /><input id="__tabbed_6_12" name="__tabbed_6" type="radio" /><div class="tabbed-labels"><label for="__tabbed_6_1">Java</label><label for="__tabbed_6_2">C++</label><label for="__tabbed_6_3">Python</label><label for="__tabbed_6_4">Go</label><label for="__tabbed_6_5">JS</label><label for="__tabbed_6_6">TS</label><label for="__tabbed_6_7">C</label><label for="__tabbed_6_8">C#</label><label for="__tabbed_6_9">Swift</label><label for="__tabbed_6_10">Zig</label><label for="__tabbed_6_11">Dart</label><label for="__tabbed_6_12">Rust</label></div>
@@ -4559,17 +4562,17 @@
 <h3 id="_5">先左旋后右旋<a class="headerlink" href="#_5" title="Permanent link">&para;</a></h3>
 <p>对于下图中的失衡节点 3，仅使用左旋或右旋都无法使子树恢复平衡。此时需要先左旋后右旋，即先对 <code>child</code> 执行「左旋」，再对 <code>node</code> 执行「右旋」。</p>
 <p><img alt="先左旋后右旋" src="../avl_tree.assets/avltree_left_right_rotate.png" /></p>
-<p align="center"> Fig. 先左旋后右旋 </p>
+<p align="center"> 图：先左旋后右旋 </p>
 
 <h3 id="_6">先右旋后左旋<a class="headerlink" href="#_6" title="Permanent link">&para;</a></h3>
 <p>同理，对于上述失衡二叉树的镜像情况，需要先右旋后左旋，即先对 <code>child</code> 执行「右旋」，然后对 <code>node</code> 执行「左旋」。</p>
 <p><img alt="先右旋后左旋" src="../avl_tree.assets/avltree_right_left_rotate.png" /></p>
-<p align="center"> Fig. 先右旋后左旋 </p>
+<p align="center"> 图：先右旋后左旋 </p>
 
 <h3 id="_7">旋转的选择<a class="headerlink" href="#_7" title="Permanent link">&para;</a></h3>
 <p>下图展示的四种失衡情况与上述案例逐个对应，分别需要采用右旋、左旋、先右后左、先左后右的旋转操作。</p>
 <p><img alt="AVL 树的四种旋转情况" src="../avl_tree.assets/avltree_rotation_cases.png" /></p>
-<p align="center"> Fig. AVL 树的四种旋转情况 </p>
+<p align="center"> 图：AVL 树的四种旋转情况 </p>
 
 <p>在代码中，我们通过判断失衡节点的平衡因子以及较高一侧子节点的平衡因子的正负号，来确定失衡节点属于上图中的哪种情况。</p>
 <div class="center-table">