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<meta name="description" content="主要是一开始没弄清题目中完全二叉树的定义 思路： 一棵排序二叉树的中序遍历就是这一组数的递增序列。这边是完全二叉树，假设从0开始，那么节点i的左孩子的标号就是2i+1，右孩子的标号就是2(i+1)。先将这组数按照递增来排序，然后用中序遍历复原这棵完全排序二叉树，最后直接输出。 完全二叉树：任何一个节点编号x，则左孩子节点为2x，右孩子节点为2x+1可以使用一个数组存储，从1到n结论：该数组中元素">

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<meta property="og:description" content="主要是一开始没弄清题目中完全二叉树的定义 思路： 一棵排序二叉树的中序遍历就是这一组数的递增序列。这边是完全二叉树，假设从0开始，那么节点i的左孩子的标号就是2i+1，右孩子的标号就是2(i+1)。先将这组数按照递增来排序，然后用中序遍历复原这棵完全排序二叉树，最后直接输出。 完全二叉树：任何一个节点编号x，则左孩子节点为2x，右孩子节点为2x+1可以使用一个数组存储，从1到n结论：该数组中元素">

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1064 Complete Binary Search Tree (30分)

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<p>主要是一开始没弄清题目中完全二叉树的定义</p>

<p>思路：</p>

<p>一棵排序二叉树的中序遍历就是这一组数的递增序列。这边是完全二叉树，假设从0开始，那么节点i的左孩子的标号就是2<em>i+1，右孩子的标号就是2</em>(i+1)。先将这组数按照递增来排序，然后用中序遍历复原这棵完全排序二叉树，最后直接输出。</p>

<p>完全二叉树：任何一个节点编号x，则左孩子节点为2x，右孩子节点为2x+1<br>可以使用一个数组存储，从1到n<br>结论：该数组中元素存放的顺序为该二叉树的层序遍历序列</p>

<p>利用完全二叉树建立结构，中序遍历填入元素</p>

<p><img src="https://i.loli.net/2020/03/04/X2rieZHWd7UNMEK.png" alt="clipboard.png"></p>

<figure class="highlight plain"><table><tr><td class="gutter"><pre><span class="line">1</span><br><span class="line">2</span><br><span class="line">3</span><br><span class="line">4</span><br><span class="line">5</span><br><span class="line">6</span><br><span class="line">7</span><br><span class="line">8</span><br><span class="line">9</span><br><span class="line">10</span><br><span class="line">11</span><br><span class="line">12</span><br><span class="line">13</span><br><span class="line">14</span><br><span class="line">15</span><br><span class="line">16</span><br><span class="line">17</span><br><span class="line">18</span><br><span class="line">19</span><br><span class="line">20</span><br><span class="line">21</span><br><span class="line">22</span><br><span class="line">23</span><br><span class="line">24</span><br><span class="line">25</span><br><span class="line">26</span><br><span class="line">27</span><br></pre></td><td class="code"><pre><span class="line">#include &lt;bits&#x2F;stdc++.h&gt;</span><br><span class="line">using namespace std;</span><br><span class="line">const int maxn &#x3D; 1000 + 10;</span><br><span class="line"></span><br><span class="line">int a[maxn], ans[maxn], cnt, N, i;</span><br><span class="line">void in_order(int root) &#123;</span><br><span class="line"> if (root &gt; N)</span><br><span class="line"> return;</span><br><span class="line"> in_order(root * 2);</span><br><span class="line"> ans[root] &#x3D; a[cnt++];</span><br><span class="line"> in_order(root * 2 + 1);</span><br><span class="line">&#125;</span><br><span class="line"></span><br><span class="line">int main(int argc, char const *argv[])</span><br><span class="line">&#123;</span><br><span class="line"> cin &gt;&gt; N;</span><br><span class="line"> for (i &#x3D; 0; i &lt; N; ++i)</span><br><span class="line"> cin &gt;&gt; a[i];</span><br><span class="line"> sort(a, a + N);</span><br><span class="line"> cnt &#x3D; 0;</span><br><span class="line"> in_order(1);</span><br><span class="line"> for (i &#x3D; 1; i &lt;&#x3D; N - 1; ++i)</span><br><span class="line"> cout &lt;&lt; ans[i] &lt;&lt; &quot; &quot;;</span><br><span class="line"> cout &lt;&lt; ans[i] &lt;&lt; endl;</span><br><span class="line"></span><br><span class="line"> return 0;</span><br><span class="line">&#125;</span><br></pre></td></tr></table></figure>

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