feat: add solutions to lcof question: 34.pathSum

添加《剑指 Offer》题解：面试题34. 二叉树中和为某一值的路径

Author: yanglbme

lcof/面试题04. 二维数组中的查找/README.md

@@ -35,8 +35,7 @@ class Solution:
         if not matrix:
             return False
 
-        i = len(matrix) - 1
-        j = 0
+        i, j = len(matrix) - 1, 0
         while i >= 0 and j < len(matrix[0]):
             if matrix[i][j] == target:
                 return True

lcof/面试题04. 二维数组中的查找/Solution.py

@@ -3,8 +3,7 @@ def findNumberIn2DArray(self, matrix: List[List[int]], target: int) -> bool:
         if not matrix:
             return False
 
-        i = len(matrix) - 1
-        j = 0
+        i, j = len(matrix) - 1, 0
         while i >= 0 and j < len(matrix[0]):
             if matrix[i][j] == target:
                 return True

lcof/面试题10- I. 斐波那契数列/README.md

@@ -37,8 +37,7 @@ class Solution:
     def fib(self, n: int) -> int:
         a, b = 0, 1
         for _ in range(n):
-            s = a + b
-            a, b = b, s
+            a, b = b, a + b
         return a % 1000000007
 ```
 

lcof/面试题10- I. 斐波那契数列/Solution.py

@@ -2,6 +2,5 @@ class Solution:
     def fib(self, n: int) -> int:
         a, b = 0, 1
         for _ in range(n):
-            s = a + b
-            a, b = b, s
+            a, b = b, a + b
         return a % 1000000007

lcof/面试题10- II. 青蛙跳台阶问题/README.md

@@ -31,8 +31,7 @@ class Solution:
     def numWays(self, n: int) -> int:
         a, b = 0, 1
         for _ in range(n):
-            s = a + b
-            a, b = b, s
+            a, b = b, a + b
         return b % 1000000007
 ```
 

lcof/面试题10- II. 青蛙跳台阶问题/Solution.py

@@ -2,6 +2,5 @@ class Solution:
     def numWays(self, n: int) -> int:
         a, b = 0, 1
         for _ in range(n):
-            s = a + b
-            a, b = b, s
-        return b % 1000000007
+            a, b = b, a + b
+        return b % 1000000007

lcof/面试题34. 二叉树中和为某一值的路径/README.md

@@ -0,0 +1,113 @@
+# [面试题34. 二叉树中和为某一值的路径](https://leetcode-cn.com/problems/er-cha-shu-zhong-he-wei-mou-yi-zhi-de-lu-jing-lcof/)
+
+## 题目描述
+<!-- 这里写题目描述 -->
+输入一棵二叉树和一个整数，打印出二叉树中节点值的和为输入整数的所有路径。从树的根节点开始往下一直到叶节点所经过的节点形成一条路径。
+
+**示例:**
+
+给定如下二叉树，以及目标和 `sum = 22`，
+
+```
+              5
+             / \
+            4   8
+           /   / \
+          11  13  4
+         /  \    / \
+        7    2  5   1
+```
+
+返回:
+
+```
+[
+   [5,4,11,2],
+   [5,8,4,5]
+]
+```
+
+**提示：**
+
+1. `节点总数 <= 10000`
+
+## 解法
+<!-- 这里可写通用的实现逻辑 -->
+先序遍历+路径记录。
+
+### Python3
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```python
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+class Solution:
+    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
+        res, path = [], []
+
+        def dfs(root, sum):
+            if not root:
+                return
+            path.append(root.val)
+            target = sum - root.val
+            if target == 0 and not (root.left or root.right):
+                res.append(list(path))
+            dfs(root.left, target)
+            dfs(root.right, target)
+            path.pop()
+        
+        dfs(root, sum)
+        return res
+
+```
+
+### Java
+<!-- 这里可写当前语言的特殊实现逻辑 -->
+
+```java
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    private List<List<Integer>> res;
+    private List<Integer> path;
+    public List<List<Integer>> pathSum(TreeNode root, int sum) {
+        res = new ArrayList<>();
+        path = new ArrayList<>();
+        dfs(root, sum);
+        return res;
+
+    }
+
+    private void dfs(TreeNode root, int sum) {
+        if (root == null) {
+            return;
+        }
+        path.add(root.val);
+        int target = sum - root.val;
+        if (target == 0 && root.left == null && root.right == null) {
+            List<Integer> t = new ArrayList<>(path);
+            res.add(t);
+        }
+        dfs(root.left, target);
+        dfs(root.right, target);
+        path.remove(path.size() - 1);
+    }
+}
+```
+
+### ...
+```
+
+```

lcof/面试题34. 二叉树中和为某一值的路径/Solution.java

@@ -0,0 +1,35 @@
+/**
+ * Definition for a binary tree node.
+ * public class TreeNode {
+ *     int val;
+ *     TreeNode left;
+ *     TreeNode right;
+ *     TreeNode(int x) { val = x; }
+ * }
+ */
+class Solution {
+    private List<List<Integer>> res;
+    private List<Integer> path;
+    public List<List<Integer>> pathSum(TreeNode root, int sum) {
+        res = new ArrayList<>();
+        path = new ArrayList<>();
+        dfs(root, sum);
+        return res;
+
+    }
+
+    private void dfs(TreeNode root, int sum) {
+        if (root == null) {
+            return;
+        }
+        path.add(root.val);
+        int target = sum - root.val;
+        if (target == 0 && root.left == null && root.right == null) {
+            List<Integer> t = new ArrayList<>(path);
+            res.add(t);
+        }
+        dfs(root.left, target);
+        dfs(root.right, target);
+        path.remove(path.size() - 1);
+    }
+}

lcof/面试题34. 二叉树中和为某一值的路径/Solution.py

@@ -0,0 +1,25 @@
+# Definition for a binary tree node.
+# class TreeNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.left = None
+#         self.right = None
+
+
+class Solution:
+    def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
+        res, path = [], []
+
+        def dfs(root, sum):
+            if not root:
+                return
+            path.append(root.val)
+            target = sum - root.val
+            if target == 0 and not (root.left or root.right):
+                res.append(list(path))
+            dfs(root.left, target)
+            dfs(root.right, target)
+            path.pop()
+
+        dfs(root, sum)
+        return res