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Author: assassint2017

Same Tree/Same_Tree.cpp

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+// 第一种思路 使用层次遍历依次比较各个节点
+// 4ms 32.81%
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+
+#include <queue>
+
+class Solution {
+public:
+    bool isSameTree(TreeNode* p, TreeNode* q) {
+        queue<TreeNode*> queue1;
+        queue<TreeNode*> queue2;
+
+        queue1.push(p);
+        queue2.push(q);
+
+        while (queue1.size() > 0 && queue2.size() > 0)
+        {
+            TreeNode *itemP = queue1.front();
+            TreeNode *itemQ = queue2.front();
+
+            queue1.pop();
+            queue2.pop();
+
+            if (itemP == 0 || itemQ == 0)
+            {
+                if (!(itemP == 0 && itemQ == 0))
+                    return false;
+            }
+            else if (!(itemP->val == itemQ->val))
+                return false;
+            else   
+            {
+                queue1.push(itemP->left);
+                queue1.push(itemP->right);
+
+                queue2.push(itemQ->left);
+                queue2.push(itemQ->right);
+            }
+        }
+
+        // 当循环退出的时候，只有两个队列都为空的时候两棵树才是完全相等的
+        if (queue1.size() == 0 && queue2.size() == 0)
+            return true;
+        else
+            return false;
+    }
+};

Same Tree/Same_Tree.py

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+# 第一种思路 使用层次遍历依次比较各个节点
+# 52ms 34.66%
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def isSameTree(self, p, q):
+        """
+        :type p: TreeNode
+        :type q: TreeNode
+        :rtype: bool
+        """
+        queue1 = [p]
+        queue2 = [q]
+        
+        while queue1 and queue2:
+            
+            itemP = queue1.pop(0)
+            itemQ = queue2.pop(0)
+            
+            if itemP is None or itemQ is None:
+                if not itemP is itemQ is None:
+                    return False
+            elif not itemP.val == itemQ.val:
+                return False
+            else:
+                queue1.append(itemP.left)
+                queue1.append(itemP.right)
+
+                queue2.append(itemQ.left)
+                queue2.append(itemQ.right)
+        
+        # 当循环退出的时候，只有两个队列都为空的时候两棵树才是完全相等的
+        return True if not queue1 and not queue2 else False