From 993b0795f5af6d9498ee98707eb780ec0c741802 Mon Sep 17 00:00:00 2001
From: ironartisan <ironerhalo@gmail.com>
Date: Thu, 5 Aug 2021 10:11:00 +0800
Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200100.=E7=9B=B8=E5=90=8C?=
 =?UTF-8?q?=E7=9A=84=E6=A0=91=20java=E5=92=8Cpython3=E7=89=88=E6=9C=AC?=
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---
 problems/0100.相同的树.md | 84 +++++++++++++++++++++++++++--------
 1 file changed, 66 insertions(+), 18 deletions(-)

diff --git a/problems/0100.相同的树.md b/problems/0100.相同的树.md
index c40f79fd..31a68692 100644
--- a/problems/0100.相同的树.md
+++ b/problems/0100.相同的树.md
@@ -139,29 +139,29 @@ public:
 ## 迭代法
 
 ```C++
-lass Solution {
+class Solution {
 public:
 
     bool isSameTree(TreeNode* p, TreeNode* q) {
         if (p == NULL && q == NULL) return true;
         if (p == NULL || q == NULL) return false;
         queue<TreeNode*> que;
-        que.push(p);   //
-        que.push(q);  //
+        que.push(p);   //  添加根节点p
+        que.push(q);  //  添加根节点q
         while (!que.empty()) {  //
             TreeNode* leftNode = que.front(); que.pop();
             TreeNode* rightNode = que.front(); que.pop();
-            if (!leftNode && !rightNode) {  //
+            if (!leftNode && !rightNode) {  // 若p的节点与q的节点都为空
                 continue;
             }
-            //
+            // 若p的节点与q的节点有一个为空或p的节点的值与q节点不同
             if ((!leftNode || !rightNode || (leftNode->val != rightNode->val))) {
                 return false;
             }
-            que.push(leftNode->left);   //
-            que.push(rightNode->left); //
-            que.push(leftNode->right);  //
-            que.push(rightNode->right);  //
+            que.push(leftNode->left);   // 添加p节点的左子树节点
+            que.push(rightNode->left); // 添加q节点的左子树节点点
+            que.push(leftNode->right);  // 添加p节点的右子树节点
+            que.push(rightNode->right);  // 添加q节点的右子树节点
         }
         return true;
     }
@@ -173,23 +173,71 @@ public:
 Java：
 
 ```java
+// 递归法
 class Solution {
     public boolean isSameTree(TreeNode p, TreeNode q) {
-        if (p == null && q == null) {
-            return true;
-        } else if (q == null || p == null) {
-            return false;
-        } else if (q.val != p.val) {
-            return false;
-        } else {
-            return isSameTree(q.left, p.left) && isSameTree(q.right, p.right);
-        }
+        if (p == null && q == null) return true;
+        else if (q == null || p == null) return false;
+        else if (q.val != p.val) return false;
+        return isSameTree(q.left, p.left) && isSameTree(q.right, p.right);
     }
 }
+```
 
+```java
+// 迭代法
+class Solution {
+    public boolean isSameTree(TreeNode p, TreeNode q) {
+        if(p == null && q == null) return true;
+        if(p == null || q == null) return false;
+        Queue<TreeNode> que= new LinkedList<TreeNode>();
+        que.offer(p);
+        que.offer(q);
+        while(!que.isEmpty()){
+            TreeNode leftNode = que.poll();
+            TreeNode rightNode = que.poll();
+            if(leftNode == null && rightNode == null) continue;
+            if(leftNode == null || rightNode== null || leftNode.val != rightNode.val) return false;
+            que.offer(leftNode.left);
+            que.offer(rightNode.left);
+            que.offer(leftNode.right);
+            que.offer(rightNode.right);
+        }
+        return true;
+    }
+}
 ```
 Python：
+```python
+# 递归法
+class Solution:
+    def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
+        if not p and not q: return True
+        elif not p or not q: return False
+        elif p.val != q.val: return False
+        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)
+```
 
+```python
+# 迭代法
+class Solution:
+    def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
+        if not p and not q: return True
+        if not p or not q: return False
+        que = collections.deque()
+        que.append(p)
+        que.append(q)
+        while que:
+            leftNode = que.popleft()
+            rightNode = que.popleft()
+            if not leftNode and not rightNode: continue 
+            if not leftNode or not rightNode or leftNode.val != rightNode.val: return False 
+            que.append(leftNode.left)
+            que.append(rightNode.left)
+            que.append(leftNode.right)
+            que.append(rightNode.right)
+        return True
+```
 Go：
 
 JavaScript：