Add some binary trees tasks

neetcode/binary_trees/insert_into_binary_tree/README.md

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+You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.
+
+Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any of them.
+
+ 
+
+Example 1:
+
+
+Input: root = [4,2,7,1,3], val = 5
+Output: [4,2,7,1,3,5]
+Explanation: Another accepted tree is:
+
+Example 2:
+
+Input: root = [40,20,60,10,30,50,70], val = 25
+Output: [40,20,60,10,30,50,70,null,null,25]
+Example 3:
+
+Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
+Output: [4,2,7,1,3,5]
+ 
+
+Constraints:
+
+The number of nodes in the tree will be in the range [0, 104].
+-108 <= Node.val <= 108
+All the values Node.val are unique.
+-108 <= val <= 108
+It's guaranteed that val does not exist in the original BST.

neetcode/binary_trees/insert_into_binary_tree/solution.py

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+from typing import Optional
+
+
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+
+class Solution:
+    def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
+        if not root:
+            return TreeNode(val, None, None)
+        if val < root.val:
+            root.left = self.insertIntoBST(root.left, val)
+        elif val > root.val:
+            root.right = self.insertIntoBST(root.right, val)
+        return root
+
+
+if __name__ == "__main__":
+    s = Solution()
+    n10 = TreeNode(1, None, None)
+    n9 = TreeNode(3, None, None)
+    n8 = TreeNode(2, n10, n9)
+    n7 = TreeNode(7, None, None)
+    root = TreeNode(4, n8, n7)
+
+    # print(s.searchBST(root, 2).val)

neetcode/binary_trees/search_in_a_binary_tree/README.md

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+700. Search in a Binary Search Tree
+You are given the root of a binary search tree (BST) and an integer val.
+
+Find the node in the BST that the node's value equals val and return the subtree rooted with that node. If such a node does not exist, return null.
+
+Example 1:
+Input: root = [4,2,7,1,3], val = 2
+Output: [2,1,3]
+
+Example 2:
+Input: root = [4,2,7,1,3], val = 5
+Output: []
+
+Constraints:
+The number of nodes in the tree is in the range [1, 5000].
+1 <= Node.val <= 107
+root is a binary search tree.
+1 <= val <= 107

neetcode/binary_trees/search_in_a_binary_tree/solution.py

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+from typing import Optional
+
+
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+
+class Solution:
+    def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
+        if not root:
+            return None
+
+        if val < root.val:
+            return self.searchBST(root.left, val)
+        elif val > root.val:
+            return self.searchBST(root.right, val)
+        else:
+            return root
+
+
+if __name__ == "__main__":
+    s = Solution()
+    n10 = TreeNode(1, None, None)
+    n9 = TreeNode(3, None, None)
+    n8 = TreeNode(2, n10, n9)
+    n7 = TreeNode(7, None, None)
+    root = TreeNode(4, n8, n7)
+
+    print(s.searchBST(root, 2).val)