feat(0783_search-in-a-binary-search-tree): add python3 soln

0783_search-in-a-binary-search-tree/README.md

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+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:**
+
+![](https://assets.leetcode.com/uploads/2021/01/12/tree1.jpg)
+
+    Input: root = [4,2,7,1,3], val = 2
+    Output: [2,1,3]
+    
+
+**Example 2:**
+
+![](https://assets.leetcode.com/uploads/2021/01/12/tree2.jpg)
+
+    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`

0783_search-in-a-binary-search-tree/python3/solution.py

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+# Definition for a binary tree node.
+# 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]:
+        current = root
+        
+        while current is not None:
+            if val == current.val:
+                return current
+            elif val < current.val:
+                current = current.left
+            else:
+                current = current.right
+        
+        return None