104

Author: wind-liang

SUMMARY.md

@@ -102,7 +102,8 @@
 * [99. Recover Binary Search Tree](leetcode-99-Recover-Binary-Search-Tree.md)
 * [100. Same Tree](leetcode-100-Same-Tree.md)
 * [leetcode 100 斩！回顾](leetcode100斩回顾.md)
-* [101 题到 103 题](leetcode-101-200.md)
+* [101 题到 104 题](leetcode-101-200.md)
     * [101. Symmetric Tree](leetcode-101-Symmetric-Tree.md)
     * [102. Binary Tree Level Order Traversal](leetcode-102-Binary-Tree-Level-Order-Traversal.md)
-    * [103. Binary Tree Zigzag Level Order Traversal](leetcode-103-Binary-Tree-Zigzag-Level-Order-Traversal.md)
+    * [103. Binary Tree Zigzag Level Order Traversal](leetcode-103-Binary-Tree-Zigzag-Level-Order-Traversal.md)
+    * [104. Maximum Depth of Binary Tree](leetcode-104-Maximum-Depth-of-Binary-Tree.md)

leetcode-101-200.md

@@ -6,3 +6,4 @@
 
 <a href="leetcode-103-Binary-Tree-Zigzag-Level-Order-Traversal.html">103. Binary Tree Zigzag Level Order Traversal</a>
 
+<a href="leetcode-104-Maximum-Depth-of-Binary-Tree.html">104. Maximum Depth of Binary Tree</a>

leetcode-103-Binary-Tree-Zigzag-Level-Order-Traversal.md

@@ -118,6 +118,8 @@ public List<List<Integer>> zigzagLevelOrder(TreeNode root) {
                 queue.offer(curNode.right);
             }
         }
+        //因为上边 queue.offer(curNode.left) 没有判断是否是 null
+        //所以要判断当前是否有元素
         if (subList.size() > 0) {
             ans.add(subList);
         }

leetcode-104-Maximum-Depth-of-Binary-Tree.md

@@ -0,0 +1,53 @@
+# 题目描述（简单难度）
+
+![](https://windliang.oss-cn-beijing.aliyuncs.com/104.jpg)
+
+输出二叉树的深度。
+
+# 解法一  DFS
+
+依旧是考的二叉树的遍历。最简单的思路就是用递归进行 DFS 即可。
+
+```java
+public int maxDepth(TreeNode root) {
+    if (root == null) {
+        return 0;
+    }
+    return Math.max(maxDepth(root.left), maxDepth(root.right)) + 1;
+}
+```
+
+# 解法二 BFS
+
+可以直接仿照 [103 题](<https://leetcode.wang/leetcode-103-Binary-Tree-Zigzag-Level-Order-Traversal.html>)，利用一个队列，进行 BFS 即可。代码可以直接搬过来。
+
+```java
+public int maxDepth(TreeNode root) {
+    Queue<TreeNode> queue = new LinkedList<TreeNode>();
+    List<List<Integer>> ans = new LinkedList<List<Integer>>();
+    if (root == null)
+        return 0;
+    queue.offer(root);
+    int level = 0;
+    while (!queue.isEmpty()) {
+        int levelNum = queue.size(); // 当前层元素的个数
+        for (int i = 0; i < levelNum; i++) {
+            TreeNode curNode = queue.poll();
+            if (curNode != null) {
+                if (curNode.left != null) {
+                    queue.offer(curNode.left);
+                }
+                if (curNode.right != null) { 
+                    queue.offer(curNode.right);
+                }
+            }
+        }
+        level++;
+    }
+    return level;
+}
+```
+
+# 总
+
+依旧考的是二叉树的遍历方式，没有什么难点。