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LeetCode 543. Diameter of Binary Tree.html
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<h2 id="543-Diameter-of-Binary-Tree（二叉树的直径）"><a href="#543-Diameter-of-Binary-Tree（二叉树的直径）" class="headerlink" title="543. Diameter of Binary Tree（二叉树的直径）"></a>543. Diameter of Binary Tree（二叉树的直径）</h2><h3 id="链接"><a href="#链接" class="headerlink" title="链接"></a>链接</h3><p><a href="https://leetcode-cn.com/problems/diameter-of-binary-tree" target="_blank" rel="noopener">https://leetcode-cn.com/problems/diameter-of-binary-tree</a></p>
<h3 id="题目"><a href="#题目" class="headerlink" title="题目"></a>题目</h3><p>给定一棵二叉树，你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过也可能不穿过根结点。</p>
<p>示例 :<br>给定二叉树</p>
<pre><code> 1
/ \
2 3
/ \
4 5 </code></pre><p>返回 3, 它的长度是路径 [4,2,1,3] 或者 [5,2,1,3]。</p>
<p>注意：两结点之间的路径长度是以它们之间边的数目表示。</p>
<h3 id="思路"><a href="#思路" class="headerlink" title="思路"></a>思路</h3><p>题目和翻译简直就是两码事，可以把题目理解成每个节点左右子树的深度和，遍历整棵树找到最大值即可，这里还是用递归舒服一点，求子树深度顺便加个max记录最大值即可。</p>
<h3 id="代码"><a href="#代码" class="headerlink" title="代码"></a>代码</h3><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></pre></td><td class="code"><pre><span class="line">public class ListNode {</span><br><span class="line"></span><br><span class="line"> int val;</span><br><span class="line"> ListNode next;</span><br><span class="line"></span><br><span class="line"> ListNode(int x) {</span><br><span class="line"> val = x;</span><br><span class="line"> }</span><br><span class="line">}</span><br><span class="line"></span><br><span class="line">public boolean isPalindrome(ListNode head) {</span><br><span class="line"> List<Integer> list = new ArrayList<Integer>();</span><br><span class="line"> while (head != null) {</span><br><span class="line"> list.add(head.val);</span><br><span class="line"> head = head.next;</span><br><span class="line"> }</span><br><span class="line"> int len = list.size();</span><br><span class="line"> Integer[] num = list.toArray(new Integer [len]);</span><br><span class="line"> for (int i = 0; i <= len / 2; i++) {</span><br><span class="line"> if (!num[i].equals(num[len - 1 - i])) {</span><br><span class="line"> return false;</span><br><span class="line"> }</span><br><span class="line"> }</span><br><span class="line"> return true;</span><br><span class="line">}</span><br></pre></td></tr></table></figure>
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