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第 5 章 &nbsp; 堆疊與佇列
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第 7 章 &nbsp; 樹
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7.1 &nbsp; 二元樹
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7.2 &nbsp; 二元樹走訪
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7.3 &nbsp; 二元樹陣列表示
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7.4 &nbsp; 二元搜尋樹
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1. &nbsp; 完全、完滿與完美二元樹
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2. &nbsp; 同一棵樹的三種走訪順序
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7.7.2 &nbsp; 程式設計練習
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1. &nbsp; 二元樹的最大深度
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2. &nbsp; 按層走訪二元樹
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第 8 章 &nbsp; 堆積
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第 9 章 &nbsp; 圖
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第 10 章 &nbsp; 搜尋
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第 11 章 &nbsp; 排序
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第 12 章 &nbsp; 分治
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12.1 &nbsp; 分治演算法
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12.3 &nbsp; 構建樹問題
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第 13 章 &nbsp; 回溯
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14.1 &nbsp; 初探動態規劃
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<h1 id="77">7.7 &nbsp; 練習<a class="headerlink" href="#77" title="Permanent link">&para;</a></h1>
<h2 id="771">7.7.1 &nbsp; 知識鞏固<a class="headerlink" href="#771" title="Permanent link">&para;</a></h2>
<h3 id="1">1. &nbsp; 完全、完滿與完美二元樹<a class="headerlink" href="#1" title="Permanent link">&para;</a></h3>
<p>下面兩個陣列按層序表示二元樹，<code>None</code> 表示空位：</p>
<ul>
<li>樹 A：<code>[1, 2, 3, 4, 5, 6]</code></li>
<li>樹 B：<code>[1, 2, 3, None, None, 6, 7]</code></li>
</ul>
<!-- numbered-subquestions -->
<ol>
<li>哪一棵是完全二元樹？</li>
<li>哪一棵是完滿二元樹，即每個非葉節點都有兩個子節點？</li>
<li>兩棵樹中是否有完美二元樹？分別說明理由。</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>樹 A 是完全二元樹。它只有最底層未填滿，並且節點從左到右連續排列。
樹 B 不是完全二元樹，因為最底層左側已有空位，右側卻仍有節點。</p>
</li>
<li>
<p>樹 B 是完滿二元樹：節點 1 和節點 3 各有兩個子節點，其餘節點都是葉節點。
樹 A 不是完滿二元樹，因為節點 3 只有左子節點 6。</p>
</li>
<li>
<p>兩棵都不是完美二元樹，因為它們的最底層都沒有完全填滿。</p>
</li>
</ol>
</details>
<h3 id="2">2. &nbsp; 同一棵樹的三種走訪順序<a class="headerlink" href="#2" title="Permanent link">&para;</a></h3>
<p>將陣列 <code>[1, 2, 3, 4, 5, 6, 7]</code> 按層序存入一棵完全二元樹。</p>
<!-- numbered-subquestions -->
<ol>
<li>畫出這棵樹。</li>
<li>寫出它的前序、中序、後序走訪序列。</li>
<li>在中序序列中，根節點 1 左側和右側的兩段序列，分別對應樹的哪一部分？</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>這棵樹為：</p>
<div class="highlight"><pre><span></span><code><a id="__codelineno-0-1" name="__codelineno-0-1" href="#__codelineno-0-1"></a> 1
<a id="__codelineno-0-2" name="__codelineno-0-2" href="#__codelineno-0-2"></a> / \
<a id="__codelineno-0-3" name="__codelineno-0-3" href="#__codelineno-0-3"></a> 2 3
<a id="__codelineno-0-4" name="__codelineno-0-4" href="#__codelineno-0-4"></a> / \ / \
<a id="__codelineno-0-5" name="__codelineno-0-5" href="#__codelineno-0-5"></a> 4 5 6 7
</code></pre></div>
</li>
<li>
<p>前序走訪為 <code>1, 2, 4, 5, 3, 6, 7</code>；
中序走訪為 <code>4, 2, 5, 1, 6, 3, 7</code>；
後序走訪為 <code>4, 5, 2, 6, 7, 3, 1</code>。</p>
</li>
<li>
<p>根節點 1 左側的 <code>4, 2, 5</code> 是左子樹的中序走訪序列；
右側的 <code>6, 3, 7</code> 是右子樹的中序走訪序列。</p>
</li>
</ol>
</details>
<h3 id="3">3. &nbsp; 比較兩棵二元搜尋樹<a class="headerlink" href="#3" title="Permanent link">&para;</a></h3>
<p>將下面兩組序列從左到右依次插入空的二元搜尋樹：</p>
<ul>
<li>序列 A：<code>[4, 2, 6, 1, 3, 5, 7]</code></li>
<li>序列 B：<code>[1, 2, 3, 4, 5, 6, 7]</code></li>
</ul>
<!-- numbered-subquestions -->
<ol>
<li>分別寫出查詢數字 7 時經過的節點。</li>
<li>若高度按根節點到最遠葉節點經過的邊數計算，兩棵樹的高度分別是多少？</li>
<li>根據前兩問，你認為兩棵樹查詢數字 7 時的效率是否相同？請結合兩棵樹的形狀和查詢路徑說明理由。</li>
</ol>
<details class="success">
<summary>參考答案</summary>
<ol>
<li>
<p>序列 A 建成的樹中，查詢路徑為 <code>4 → 6 → 7</code>；
序列 B 建成的樹中，查詢路徑為 <code>1 → 2 → 3 → 4 → 5 → 6 → 7</code>。</p>
</li>
<li>
<p>第一棵樹的每一層都填滿，高度為 2；第二棵樹只有右孩子，高度為 6。</p>
</li>
<li>
<p>兩棵樹查詢 7 時的效率不同。插入順序改變了二元搜尋樹的形狀和高度。查詢 7 時，第一棵樹只經過 3 個節點，
第二棵樹則要依次經過 7 個節點；樹越高，最壞情況下需要沿路徑比較的節點就越多。</p>
</li>
</ol>
</details>
<h2 id="772">7.7.2 &nbsp; 程式設計練習<a class="headerlink" href="#772" title="Permanent link">&para;</a></h2>
<h3 id="1_1">1. &nbsp; 二元樹的最大深度<a class="headerlink" href="#1_1" title="Permanent link">&para;</a></h3>
<p>給定一棵二元樹的根節點 <code>root</code>。每個節點都包含一個整數值，以及指向左、右子節點的引用。</p>
<p>把從根節點到最遠葉節點經過的**節點數**稱為二元樹的最大深度。請返回這棵樹的最大深度；空樹的最大深度為 0。
請使用遞迴完成。</p>
<details class="tip">
<summary>解題提示</summary>
<ol>
<li>本題按節點數計算深度：只有一個根節點時，最大深度為 1</li>
<li>讓遞迴函式返回以當前節點為根的子樹的最大深度</li>
<li>空節點返回 0，非空節點返回 max(depth(left), depth(right)) + 1</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/maximum-depth-of-binary-tree/" rel="noopener noreferrer" target="_blank">LeetCode</a> <a class="rounded-button exercise-button" href="https://leetcode.cn/problems/maximum-depth-of-binary-tree/solutions/2361697/104-er-cha-shu-de-zui-da-shen-du-hou-xu-txzrx/" rel="noopener noreferrer" target="_blank">題目解析</a></p>
<h3 id="2_1">2. &nbsp; 按層走訪二元樹<a class="headerlink" href="#2_1" title="Permanent link">&para;</a></h3>
<p>給定一棵二元樹的根節點 <code>root</code>。請使用佇列從上到下逐層訪問所有節點，同一層內按照從左到右的順序訪問。</p>
<p>返回一個二維陣列：第一個子陣列儲存根節點所在層的節點值，第二個子陣列儲存下一層的節點值，依此類推。
若二元樹為空，返回空陣列。</p>
<details class="tip">
<summary>解題提示</summary>
<ol>
<li>層序走訪需要先進入的節點先訪問，因此使用佇列</li>
<li>一輪開始時佇列中的節點正好屬於同一層</li>
<li>先記錄佇列長度，再彈出這麼多個節點並加入它們的孩子</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/binary-tree-level-order-traversal/" rel="noopener noreferrer" target="_blank">LeetCode</a> <a class="rounded-button exercise-button" href="https://leetcode.cn/problems/binary-tree-level-order-traversal/solutions/2361604/102-er-cha-shu-de-ceng-xu-bian-li-yan-du-dyf7/" rel="noopener noreferrer" target="_blank">題目解析</a></p>
<h3 id="3-k">3. &nbsp; 二元搜尋樹中的第 k 小元素<a class="headerlink" href="#3-k" title="Permanent link">&para;</a></h3>
<p>一棵二元搜尋樹共有 <code>n</code> 個節點，各節點值互不相同。
將所有節點值從小到大排列後，位置從 1 開始編號。</p>
<p>給定根節點 <code>root</code> 和滿足 <code>1 &lt;= k &lt;= n</code> 的整數 <code>k</code>，請返回排在第 <code>k</code> 位的節點值。
請在中序走訪過程中直接尋找答案，不先收集全部節點值。</p>
<details class="tip">
<summary>解題提示</summary>
<ol>
<li>二元搜尋樹的中序走訪會按從小到大的順序訪問節點值</li>
<li>中序走訪依次處理左子樹、當前節點和右子樹；訪問當前節點時把計數加 1</li>
<li>當計數第一次等於 k 時，當前節點值就是答案，此後無須繼續走訪</li>
</ol>
</details>
<p><a class="rounded-button exercise-button" href="https://leetcode.cn/problems/kth-smallest-element-in-a-bst/" rel="noopener noreferrer" target="_blank">LeetCode</a> <a class="rounded-button exercise-button" href="https://leetcode.cn/problems/kth-smallest-element-in-a-bst/solutions/2361685/230-er-cha-sou-suo-shu-zhong-di-k-xiao-d-n3he/" rel="noopener noreferrer" target="_blank">題目解析</a></p>
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