added 0094_binary_tree_inorder_traversal.py

Author: nyierr

problems/easy/0094_binary_tree_inorder_traversal.py

@@ -0,0 +1,91 @@
+"""
+Given the root of a binary tree, return the inorder traversal of its
+nodes' values.
+
+Example 1:
+  Input: root = [1,null,2,3]
+  Output: [1,3,2]
+    
+Example 2:
+  Input: root = []
+  Output: []
+
+Example 3:
+  Input: root = [1]
+  Output: [1]
+
+Constraints:
+  * The number of nodes in the tree is in the range [0, 100]
+  * -100 <= Node.val <= 100
+  
+Follow up: Recursive solution is trivial, could you do it iteratively?
+"""
+
+class Solution:          
+    # O(2^n) recursive solution, using default arg to collect result
+    #def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
+    #    res = []
+    #    self.helper(root, res)
+    #    return res
+    #
+    #def helper(self, node, res=[]):
+    #    if not node is None:
+    #        self.helper(node.left, res)
+    #        res.append(node.val)
+    #        self.helper(node.right, res)
+    
+    # O(2^n) recursive solution
+    #def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
+    #    if root is None:
+    #        return [] 
+    #    res = self.inorderTraversal(root.left)
+    #    res.append(root.val)
+    #    res += self.inorderTraversal(root.right)
+    #    return res
+    
+    # O(2^n) iterative solution (breadth-first)
+    #def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
+    #    if root is None:
+    #        return []
+    #    
+    #    to_do = [root.right, root, root.left]
+    #    visited = {root: True}
+    #    
+    #    res = []
+    #    while len(to_do):
+    #        curr = to_do[-1]
+    #        if curr is None:
+    #            to_do.pop()
+    #            continue
+    #        
+    #        # Already visited or a leaf
+    #        if curr in visited or (curr.left is None and curr.right is None):
+    #            to_do.pop()
+    #            res.append(curr.val)
+    #            continue
+    #
+    #        if not curr.right is None:
+    #            to_do.insert(-1, curr.right)
+    #        if not curr.left is None:
+    #            to_do.append(curr.left)
+    #            
+    #        visited[curr] = True                
+    #        
+    #    return res
+    
+    # O(2^n) iterative solution (depth-first)
+    def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
+        res = []
+        stack = []
+        
+        while True:
+            while root:
+                stack.append(root)
+                root = root.left
+                
+            if len(stack) == 0:
+                return res
+            
+            curr = stack.pop()
+            res.append(curr.val)
+            root = curr.right