Merge pull request #449 from KailokFung/master

feat(二叉树理论基础.md): 新增java版本

problems/0226.翻转二叉树.md

@@ -205,6 +205,7 @@ public:
 Java：
 
 ```Java
+//DFS递归
 class Solution {
    /**
      * 前后序遍历都可以
@@ -226,6 +227,31 @@ class Solution {
         root.right = tmp;
     }
 }
+
+//BFS
+class Solution {
+    public TreeNode invertTree(TreeNode root) {
+        if (root == null) {return null;}
+        ArrayDeque<TreeNode> deque = new ArrayDeque<>();
+        deque.offer(root);
+        while (!deque.isEmpty()) {
+            int size = deque.size();
+            while (size-- > 0) {
+                TreeNode node = deque.poll();
+                swap(node);
+                if (node.left != null) {deque.offer(node.left);}
+                if (node.right != null) {deque.offer(node.right);}
+            }
+        }
+        return root;
+    }
+
+    public void swap(TreeNode root) {
+        TreeNode temp = root.left;
+        root.left = root.right;
+        root.right = temp;
+    }
+}
 ```
 
 Python：

problems/0257.二叉树的所有路径.md

@@ -283,6 +283,7 @@ public:
 Java：
 
 ```Java
+//解法一
 class Solution {
     /**
      * 递归法
@@ -321,6 +322,52 @@ class Solution {
     }
 }
 
+//解法二（常规前序遍历，不用回溯），更容易理解
+class Solution {
+    public List<String> binaryTreePaths(TreeNode root) {
+        List<String> res = new ArrayList<>();
+        helper(root, new StringBuilder(), res);
+        return res;
+    }
+
+    public void helper(TreeNode root, StringBuilder sb, List<String> res) {
+        if (root == null) {return;}
+      	// 遇到叶子结点就放入当前路径到res集合中
+        if (root.left == null && root.right ==null) {
+            sb.append(root.val);
+            res.add(sb.toString());
+          	// 记得结束当前方法
+            return;
+        }
+        helper(root.left,new StringBuilder(sb).append(root.val + "->"),res);
+        helper(root.right,new StringBuilder(sb).append(root.val + "->"),res);
+    }
+}
+
+//针对解法二优化，思路本质是一样的
+class Solution {
+    public List<String> binaryTreePaths(TreeNode root) {
+        List<String> res = new ArrayList<>();
+        helper(root, "", res);
+        return res;
+    }
+
+    public void helper(TreeNode root, String path, List<String> res) {
+        if (root == null) {return;}
+      	// 由原始解法二可以知道，root的值肯定会下面某一个条件加入到path中，那么干脆直接在这一步加入即可
+        StringBuilder sb = new StringBuilder(path);
+        sb.append(root.val);
+        if (root.left == null && root.right ==null) {
+            res.add(sb.toString());
+        }else{
+          	// 如果是非叶子结点则还需要跟上一个 “->”
+            sb.append("->");
+            helper(root.left,sb.toString(),res);
+            helper(root.right,sb.toString(),res);
+        }
+    }
+}
+
 ```
 
 Python：
@@ -350,7 +397,7 @@ class Solution:
 
 ```
 Go：
-  
+
 ```go
 func binaryTreePaths(root *TreeNode) []string {
     res := make([]string, 0)

problems/二叉树理论基础.md

@@ -188,6 +188,21 @@ struct TreeNode {
 
 Java：
 
+```java
+public class TreeNode {
+    int val;
+  	TreeNode left;
+  	TreeNode right;
+  	TreeNode() {}
+  	TreeNode(int val) { this.val = val; }
+  	TreeNode(int val, TreeNode left, TreeNode right) {
+    		this.val = val;
+    		this.left = left;
+    		this.right = right;
+  	}
+}
+```
+
 
 Python：