upload solutions 144 & 145

Author: BrianRuizy

144 - Binary Tree Preorder Traversal.md

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+# 144 - Binary Tree Preorder Traversal
+
+Difficulty: easy
+Done: Yes
+Last edited: March 13, 2022 10:59 PM
+Link: https://leetcode.com/problems/binary-tree-preorder-traversal/
+Topic: recursion, tree
+
+## Problem
+
+---
+
+Given the `root` of a binary tree, return *the preorder traversal of its nodes' values*.
+
+```
+Input: root = [1,null,2,3]
+Output: [1,2,3]
+```
+
+## Solution
+
+---
+
+Pre-order traversal requires order of parent -> left child -> right child. Can do so recursively, and return empty array if input is none
+
+## Code
+
+---
+
+```python
+# 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 preorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
+        
+        # if we have no nodes in input return empty list
+        if not root: 
+            return []
+        
+        # pre-order traversal requires order of parent -> left child -> right child
+        # can do so recursively 
+        
+        return [root.val] + self.preorderTraversal(root.left) + self.preorderTraversal(root.right)
+```
+
+## Time Complexity
+
+---
+
+The time complexity is O(n)*O*(*n*) because the recursive function is $T(n)=2⋅T(n/2)+1T(n)=2⋅T(n/2)+1$.

145 - Binary Tree Postorder Traversal.md

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+# 145 - Binary Tree Postorder Traversal
+
+Difficulty: easy
+Done: Yes
+Last edited: March 13, 2022 10:59 PM
+Link: https://leetcode.com/problems/binary-tree-postorder-traversal/
+Topic: recursion, tree
+
+## Problem
+
+---
+
+Given the `root` of a binary tree, return *the preorder traversal of its nodes' values*.
+
+```
+Input: root = [1,null,2,3]
+Output: [3,2,1]
+```
+
+## Solution
+
+---
+
+post order goes as left child -> right child -> root
+
+## Code
+
+---
+
+```python
+# 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 postorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
+        return self.postorderTraversal(root.left) + self.postorderTraversal(root.right) + [root.val] if  root is not None else []
+```
+
+## Time Complexity
+
+---
+
+The time complexity is O(n)*O*(*n*) because the recursive function is $T(n)=2⋅T(n/2)+1T(n)=2⋅T(n/2)+1$.