Update 二叉树中递归带着回溯.md

problems/二叉树中递归带着回溯.md

@@ -257,6 +257,83 @@ Java：
 
 Python：
 
+100.相同的树
+> 递归法
+```python
+class Solution:
+    def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
+        return self.compare(p, q)
+        
+    def compare(self, tree1, tree2):
+        if not tree1 and tree2:
+            return False
+        elif tree1 and not tree2:
+            return False
+        elif not tree1 and not tree2:
+            return True
+        elif tree1.val != tree2.val: #注意这里我没有使用else
+            return False
+        
+        #此时就是：左右节点都不为空，且数值相同的情况
+        #此时才做递归，做下一层的判断
+        compareLeft = self.compare(tree1.left, tree2.left) #左子树：左、 右子树：左
+        compareRight = self.compare(tree1.right, tree2.right) #左子树：右、 右子树：右
+        isSame = compareLeft and compareRight #左子树：中、 右子树：中（逻辑处理）
+        return isSame
+```
+
+257.二叉的所有路径
+> 递归中隐藏着回溯
+```python
+class Solution:
+    def binaryTreePaths(self, root: TreeNode) -> List[str]:
+        result = []
+        path = []
+        if not root:
+            return result
+        self.traversal(root, path, result)
+        return result
+        
+    def traversal(self, cur, path, result):
+        path.append(cur.val)
+        #这才到了叶子节点
+        if not cur.left and not cur.right:
+            sPath = ""
+            for i in range(len(path)-1):
+                sPath += str(path[i])
+                sPath += "->"
+            sPath += str(path[len(path)-1])
+            result.append(sPath)
+            return
+        if cur.left:
+            self.traversal(cur.left, path, result)
+            path.pop() #回溯
+        if cur.right:
+            self.traversal(cur.right, path, result)
+            path.pop() #回溯
+```
+
+> 精简版
+```python
+class Solution:
+    def binaryTreePaths(self, root: TreeNode) -> List[str]:
+        result = []
+        path = ""
+        if not root:
+            return result
+        self.traversal(root, path, result)
+        return result
+        
+    def traversal(self, cur, path, result):
+        path += str(cur.val) #中
+        if not cur.left and not cur.right:
+            result.append(path)
+            return
+        if cur.left:
+            self.traversal(cur.left, path+"->", result) #左  回溯就隐藏在这里
+        if cur.right:
+            self.traversal(cur.right, path+"->", result) #右 回溯就隐藏在这里
+```
 
 Go：