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wind-liang · Jul 20, 2019 · 0d4c1c5 · 0d4c1c5
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diff --git a/SUMMARY.md b/SUMMARY.md
@@ -102,10 +102,11 @@
 * [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 106 题](leetcode-101-200.md)
+* [101 题到 107 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)
     * [104. Maximum Depth of Binary Tree](leetcode-104-Maximum-Depth-of-Binary-Tree.md)
     * [105. Construct Binary Tree from Preorder and Inorder Traversal](leetcode-105-Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.md)
-    * [106. Construct Binary Tree from Inorder and Postorder Traversal](leetcode-106-Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md)
+    * [106. Construct Binary Tree from Inorder and Postorder Traversal](leetcode-106-Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.md)
+    * [107. Binary Tree Level Order Traversal II](leetcode-107-Binary-Tree-Level-Order-TraversalII.md)
diff --git a/leetcode-101-200.md b/leetcode-101-200.md
@@ -8,4 +8,6 @@
 
 <a href="leetcode-105-Construct-Binary-Tree-from-Preorder-and-Inorder-Traversal.html">105. Construct Binary Tree from Preorder and Inorder Traversal</a>
 
-<a href="leetcode-106-Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.html">106. Construct Binary Tree from Inorder and Postorder Traversal</a>
+<a href="leetcode-106-Construct-Binary-Tree-from-Inorder-and-Postorder-Traversal.html">106. Construct Binary Tree from Inorder and Postorder Traversal</a>
+
+<a href="leetcode-107-Binary-Tree-Level-Order-TraversalII.html">107. Binary Tree Level Order Traversal II</a>
diff --git a/leetcode-107-Binary-Tree-Level-Order-TraversalII.md b/leetcode-107-Binary-Tree-Level-Order-TraversalII.md
@@ -0,0 +1,173 @@
+# 题目描述（简单难度）
+
+![](https://windliang.oss-cn-beijing.aliyuncs.com/107.jpg)
+
+树的层次遍历，和 [102 题](<https://leetcode.wang/leetcode-102-Binary-Tree-Level-Order-Traversal.html>) 的不同之处是，之前输出的数组顺序是从根部一层一层的输出，现在是从底部，一层一层的输出。
+
+# 解法一 DFS
+
+把 [102 题](<https://leetcode.wang/leetcode-102-Binary-Tree-Level-Order-Traversal.html>) 的`DFS`贴过来看一下。
+
+```java
+public List<List<Integer>> levelOrder(TreeNode root) {
+    List<List<Integer>> ans = new ArrayList<>(); 
+    DFS(root, 0, ans);
+    return ans;
+}
+
+private void DFS(TreeNode root, int level, List<List<Integer>> ans) {
+    if(root == null){
+        return;
+    }
+    //当前层数还没有元素，先 new 一个空的列表
+    if(ans.size()<=level){
+        ans.add(new ArrayList<>());
+    }
+    //当前值加入
+    ans.get(level).add(root.val);
+
+    DFS(root.left,level+1,ans);
+    DFS(root.right,level+1,ans);
+} 
+```
+
+之前我们根据 level 得到数组的位置，然后添加。
+
+```java
+ans.get(level).add(root.val);
+
+ans    [] [] [] [] [].
+index  0  1  2  3  4
+level  0  1  2  3  4
+       ------------>
+index = 0 + level
+
+现在 level 是逆过来存的
+ans    [] [] [] [] [].
+index  0  1  2  3  4
+level  4  3  2  1  0
+       <------------
+index = 4 - level
+
+4 就是 ans 的末尾下标，就是 ans.size() - 1
+所以代码变为
+ans.get(ans.size() - 1 - level).add(root.val);
+```
+
+此外还有句代码要改。
+
+```java
+if(ans.size()<=level){
+    ans.add(new ArrayList<>());
+}
+在添加当前 level 的第一个元素的时候，首先添加一个空列表到 ans 中
+假设当前 level = 2，ans 中只添加了 level 是 0 和 1 的元素
+ans    [3] [9] 
+index   0   1  
+level   1   0  
+因为 level 是从右往左增加的，所以空列表要到 ans 的头部
+ans     [] [3] [9] 
+index   0   1   2
+level   2   1   0  
+所以代码改成下边的样子
+ ans.add(0，new ArrayList<>());
+```
+
+综上，只要改了这两处就可以了。
+
+```java
+public List<List<Integer>> levelOrderBottom(TreeNode root) {
+    List<List<Integer>> ans = new ArrayList<>();
+    DFS(root, 0, ans);
+    return ans;
+}
+
+private void DFS(TreeNode root, int level, List<List<Integer>> ans) {
+    if (root == null) {
+        return;
+    }
+    // 当前层数还没有元素，先 new 一个空的列表
+    if (ans.size() <= level) {
+        ans.add(0, new ArrayList<>());
+    }
+    // 当前值加入
+    ans.get(ans.size() - 1 - level).add(root.val);
+
+    DFS(root.left, level + 1, ans);
+    DFS(root.right, level + 1, ans);
+}
+```
+
+# 解法二 BFS
+
+ [102 题](<https://leetcode.wang/leetcode-102-Binary-Tree-Level-Order-Traversal.html>)  从根节点往下走的代码贴过来。
+
+```java
+public List<List<Integer>> levelOrder(TreeNode root) {
+    Queue<TreeNode> queue = new LinkedList<TreeNode>();
+    List<List<Integer>> ans = new LinkedList<List<Integer>>();
+    if (root == null)
+        return ans;
+    queue.offer(root);
+    while (!queue.isEmpty()) {
+        int levelNum = queue.size(); // 当前层元素的个数
+        List<Integer> subList = new LinkedList<Integer>();
+        for (int i = 0; i < levelNum; i++) {
+            TreeNode curNode = queue.poll();
+            if (curNode != null) {
+                subList.add(curNode.val); 
+                queue.offer(curNode.left);
+                queue.offer(curNode.right);
+            }
+        }
+        if(subList.size()>0){
+            ans.add(subList);
+        }
+    }
+    return ans;
+} 
+```
+
+`BFS`相比于`DFS`要简单些，因为`BFS`是一次性把当前层的元素都添加到`ans`中，所以我们只需要改一句代码。
+
+```java
+ans.add(subList);
+```
+
+改成添加到头部即可。
+
+```java
+ans.add(0,subList);
+```
+
+再改个函数名字， 总体代码就是
+
+```java
+public List<List<Integer>> levelOrderBottom(TreeNode root) {
+    Queue<TreeNode> queue = new LinkedList<TreeNode>();
+    List<List<Integer>> ans = new LinkedList<List<Integer>>();
+    if (root == null)
+        return ans;
+    queue.offer(root);
+    while (!queue.isEmpty()) {
+        int levelNum = queue.size(); // 当前层元素的个数
+        List<Integer> subList = new LinkedList<Integer>();
+        for (int i = 0; i < levelNum; i++) {
+            TreeNode curNode = queue.poll();
+            if (curNode != null) {
+                subList.add(curNode.val);
+                queue.offer(curNode.left);
+                queue.offer(curNode.right);
+            }
+        }
+        if (subList.size() > 0) {
+            ans.add(0, subList);
+        }
+    }
+    return ans;
+}
+```
+
+# 总
+
+这道题依旧考层次遍历，只需要在  [102 题](<https://leetcode.wang/leetcode-102-Binary-Tree-Level-Order-Traversal.html>) 的基础上，找到 `level` 和 `index` 的对应关系即可。