feat: add python and java solutions to lcci problem: No.04.10 and No.04.12

lcci/04.10.Check SubTree/README.md

@@ -29,20 +29,54 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-
+先找 t1 中 t2 结点,找到后进行 DFS，确认子树和 t2 的子树完全相同，否则返回 FALSE。
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+class Solution:
+    def checkSubTree(self, t1: TreeNode, t2: TreeNode) -> bool:
+        if t1 == None:
+            return False
+        if t2 == None:
+            return True
+        return self.dfs(t1,t2) or self.checkSubTree(t1.left,t2) or self.checkSubTree(t1.right,t2)
+
+    def dfs(self, t1: TreeNode, t2: TreeNode) -> bool:
+        if not t1 and t2 :
+            return False
+        if not t2 and not t1:
+            return True
+        if t1.val != t2.val:
+            return False
+        else:
+            return self.dfs(t1.left,t2.left) and self.dfs(t1.right,t2.right)
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+class Solution {
+    public boolean checkSubTree(TreeNode t1, TreeNode t2) {
+        if (t2 == null)
+            return true;
+        if (t1 == null)
+            return false;
+        return isSubTree(t1, t2) || checkSubTree(t1.left, t2) || checkSubTree(t1.right, t2);
+    }
+
+    public boolean isSubTree(TreeNode t1, TreeNode t2){
+        if (t2 == null)
+            return true;
+        if (t1 == null)
+            return false;
+        if (t1.val != t2.val)
+            return false;
+        return isSubTree(t1.left,t2.left) && isSubTree(t1.right,t2.right);
+    }
+}
 ```
 
 ### ...

lcci/04.10.Check SubTree/README_EN.md

@@ -51,18 +51,52 @@
 
 
 ## Solutions
-
+Find the t2 node in t1 first, then use the depth-first search (DFS) algorithm to make sure that the subtree and the subtree of t2 are identical, otherwise return FALSE.
 
 ### Python3
 
 ```python
-
+class Solution:
+    def checkSubTree(self, t1: TreeNode, t2: TreeNode) -> bool:
+        if t1 == None:
+            return False
+        if t2 == None:
+            return True
+        return self.dfs(t1,t2) or self.checkSubTree(t1.left,t2) or self.checkSubTree(t1.right,t2)
+
+    def dfs(self, t1: TreeNode, t2: TreeNode) -> bool:
+        if not t1 and t2 :
+            return False
+        if not t2 and not t1:
+            return True
+        if t1.val != t2.val:
+            return False
+        else:
+            return self.dfs(t1.left,t2.left) and self.dfs(t1.right,t2.right)
 ```
 
 ### Java
 
 ```java
-
+class Solution {
+    public boolean checkSubTree(TreeNode t1, TreeNode t2) {
+        if (t2 == null)
+            return true;
+        if (t1 == null)
+            return false;
+        return isSubTree(t1, t2) || checkSubTree(t1.left, t2) || checkSubTree(t1.right, t2);
+    }
+
+    public boolean isSubTree(TreeNode t1, TreeNode t2){
+        if (t2 == null)
+            return true;
+        if (t1 == null)
+            return false;
+        if (t1.val != t2.val)
+            return false;
+        return isSubTree(t1.left,t2.left) && isSubTree(t1.right,t2.right);
+    }
+}
 ```
 
 ### ...

lcci/04.10.Check SubTree/Solution.java

@@ -0,0 +1,19 @@
+class Solution {
+    public boolean checkSubTree(TreeNode t1, TreeNode t2) {
+        if (t2 == null)
+            return true;
+        if (t1 == null)
+            return false;
+        return isSubTree(t1, t2) || checkSubTree(t1.left, t2) || checkSubTree(t1.right, t2);
+    }
+
+    public boolean isSubTree(TreeNode t1, TreeNode t2){
+        if (t2 == null)
+            return true;
+        if (t1 == null)
+            return false;
+        if (t1.val != t2.val)
+            return false;
+        return isSubTree(t1.left,t2.left) && isSubTree(t1.right,t2.right);
+    }
+}

lcci/04.10.Check SubTree/Solution.py

@@ -0,0 +1,17 @@
+class Solution:
+    def checkSubTree(self, t1: TreeNode, t2: TreeNode) -> bool:
+        if t1 == None:
+            return False
+        if t2 == None:
+            return True
+        return self.dfs(t1,t2) or self.checkSubTree(t1.left,t2) or self.checkSubTree(t1.right,t2)
+
+    def dfs(self, t1: TreeNode, t2: TreeNode) -> bool:
+        if not t1 and t2 :
+            return False
+        if not t2 and not t1:
+            return True
+        if t1.val != t2.val:
+            return False
+        else:
+            return self.dfs(t1.left,t2.left) and self.dfs(t1.right,t2.right)

lcci/04.12.Paths with Sum/README.md

@@ -32,20 +32,73 @@
 
 ## 解法
 <!-- 这里可写通用的实现逻辑 -->
-
+DFS 深度优先搜索
 
 ### Python3
 <!-- 这里可写当前语言的特殊实现逻辑 -->
+采用递归的思想，每递归到某个节点时：
+- 若`root.val-sum == 0`，结果加 1
+- 考虑将此节点纳入或不纳入路径两种情况
 
-```python
+特殊情况：若此节点的父节点在路径中，此节点必纳入路径（路径不能断）
 
+```python
+class Solution:
+    def pathSum(self, root: TreeNode, sum: int) -> int:
+        def dfs(root, sum, flag):
+            nonlocal ans
+            if not root:
+                return 0
+            if sum-root.val == 0:
+                ans += 1
+            if flag == 0:
+                dfs(root.left, sum, 0)
+                dfs(root.right, sum, 0)
+            dfs(root.left, sum-root.val, 1)
+            dfs(root.right, sum-root.val, 1)
+
+        if not root:
+            return 0
+        ans = 0
+        dfs(root, sum, 0)
+        return ans
 ```
 
 ### Java
 <!-- 这里可写当前语言的特殊实现逻辑 -->
+使用到2个递归过程：
 
-```java
+- BFS:(traverse)遍历每个树节点；
+- DFS: 从每个树节点出发，节点求和，看是否能满足 sum。
+
+需要注意，节点值有正有负，需要穷尽所有的可能路径。
 
+```java
+class Solution {
+    int ans = 0;
+    public int pathSum(TreeNode root, int sum) {
+        traverse(root, sum);
+        return ans;
+    }
+
+    void traverse(TreeNode root, int sum) {
+        if (root == null) return;
+        ans += dfs(root, sum, 0);
+        traverse(root.left,  sum);
+        traverse(root.right, sum);
+    }
+
+    // check if sum of path is sum.
+    int dfs(TreeNode root, int sum, int cur) {
+        if (root == null) return 0;
+        cur += root.val;
+        int res = 0;
+        if (cur == sum) res++;
+        res += dfs(root.left,  sum, cur);
+        res += dfs(root.right, sum, cur);
+        return res;
+    }
+}
 ```
 
 ### ...

lcci/04.12.Paths with Sum/README_EN.md

@@ -57,18 +57,71 @@ Given the following tree and &nbsp;<code>sum = 22,</code></p>
 
 
 ## Solutions
-
+Depth-First-Search
 
 ### Python3
+Using the idea of recursion, at each recursion to a node.
+- If root.val-sum == 0, add 1 to the result
+- Consider two scenarios for inclusion or exclusion of this node from the pathway
 
-```python
+Special case: if the parent node of this node is in the path, this node must be included in the path (the path cannot be broken)
 
+```python
+class Solution:
+    def pathSum(self, root: TreeNode, sum: int) -> int:
+        def dfs(root, sum, flag):
+            nonlocal ans
+            if not root:
+                return 0
+            if sum-root.val == 0:
+                ans += 1
+            if flag == 0:
+                dfs(root.left, sum, 0)
+                dfs(root.right, sum, 0)
+            dfs(root.left, sum-root.val, 1)
+            dfs(root.right, sum-root.val, 1)
+
+        if not root:
+            return 0
+        ans = 0
+        dfs(root, sum, 0)
+        return ans
 ```
 
 ### Java
+Use to 2 recursive processes.
 
-```java
+- BFS: (traverse) traverses each tree node.
+- DFS: Starting from each tree node, the nodes sum to see if sum can be satisfied.
+
+Note that node values can be positive or negative, and all possible paths need to be exhausted.
 
+```java
+class Solution {
+    int ans = 0;
+    public int pathSum(TreeNode root, int sum) {
+        traverse(root, sum);
+        return ans;
+    }
+
+    void traverse(TreeNode root, int sum) {
+        if (root == null) return;
+        ans += dfs(root, sum, 0);
+        traverse(root.left,  sum);
+        traverse(root.right, sum);
+    }
+
+    // check if sum of path is sum.
+    int dfs(TreeNode root, int sum, int cur) {
+        if (root == null) return 0;
+        cur += root.val;
+        int res = 0;
+        if (cur == sum) res++;
+        res += dfs(root.left,  sum, cur);
+        res += dfs(root.right, sum, cur);
+        return res;
+    }
+}
 ```
 
 ### ...

lcci/04.12.Paths with Sum/Solution.java

@@ -0,0 +1,25 @@
+class Solution {
+    int ans = 0;
+    public int pathSum(TreeNode root, int sum) {
+        traverse(root, sum);
+        return ans;
+    }
+
+    void traverse(TreeNode root, int sum) {
+        if (root == null) return;
+        ans += dfs(root, sum, 0);
+        traverse(root.left,  sum);
+        traverse(root.right, sum);
+    }
+
+    // check if sum of path is sum.
+    int dfs(TreeNode root, int sum, int cur) {
+        if (root == null) return 0;
+        cur += root.val;
+        int res = 0;
+        if (cur == sum) res++;
+        res += dfs(root.left,  sum, cur);
+        res += dfs(root.right, sum, cur);
+        return res;
+    }
+}

lcci/04.12.Paths with Sum/Solution.py

@@ -0,0 +1,19 @@
+class Solution:
+    def pathSum(self, root: TreeNode, sum: int) -> int:
+        def dfs(root, sum, flag):
+            nonlocal ans
+            if not root:
+                return 0
+            if sum-root.val == 0:
+                ans += 1
+            if flag == 0:
+                dfs(root.left, sum, 0)
+                dfs(root.right, sum, 0)
+            dfs(root.left, sum-root.val, 1)
+            dfs(root.right, sum-root.val, 1)
+
+        if not root:
+            return 0
+        ans = 0
+        dfs(root, sum, 0)
+        return ans