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Author: assassint2017

Path Sum/Path_Sum.cpp

@@ -0,0 +1,33 @@
+// 第一种思路 使用递归
+// 8ms 57.10%
+
+/**
+ * Definition for a binary tree node.
+ * struct TreeNode {
+ *     int val;
+ *     TreeNode *left;
+ *     TreeNode *right;
+ *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
+ * };
+ */
+
+class Solution {
+public:
+    bool hasPathSum(TreeNode* root, int sum) {
+        
+        // 如果是空树，则直接返回
+        if (root == 0)
+            return false;
+
+        // 如果查询到了叶子节点
+        if (root->left == 0 && root->right == 0)
+        {
+            if (sum == root->val)
+                return true;
+            else
+                return false;
+        }
+            
+       return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
+    }
+};

Path Sum/Path_Sum.py

@@ -0,0 +1,30 @@
+# 第一种思路 使用递归
+# 76ms 39.18%
+
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def hasPathSum(self, root, sum):
+        """
+        :type root: TreeNode
+        :type sum: int
+        :rtype: bool
+        """
+        # 如果是空树，则直接返回
+        if root is None:
+            return False
+        
+        # 如果查询到了叶子节点
+        if root.left is None and root.right is None:
+            if sum == root.val:
+                return True
+            else:
+                return False
+        
+        return self.hasPathSum(root.left, sum-root.val) or self.hasPathSum(root.right, sum-root.val)
+