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Author: NamVr

0094-binary-tree-inorder-traversal/README.md

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+<p>Given the <code>root</code> of a binary tree, return <em>the inorder traversal of its nodes&#39; values</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = [1,null,2,3]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[1,3,2]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/screenshot-2024-08-29-202743.png" style="width: 200px; height: 264px;" /></p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = [1,2,3,4,5,null,8,null,null,6,7,9]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[4,2,6,5,7,1,3,9,8]</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/29/tree_2.png" style="width: 350px; height: 286px;" /></p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = []</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[]</span></p>
+</div>
+
+<p><strong class="example">Example 4:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">root = [1]</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">[1]</span></p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the tree is in the range <code>[0, 100]</code>.</li>
+	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
+</ul>
+
+<p>&nbsp;</p>
+<strong>Follow up:</strong> Recursive solution is trivial, could you do it iteratively?

0094-binary-tree-inorder-traversal/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 inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
+        # LEFT -> ROOT -> RIGHT
+        # return self.inorderTraversal(root.left) + [root.val] + self.inorderTraversal(root.right) if root else []
+
+        stack = []
+        res = []
+
+        while root or stack:
+            while root:
+                stack.append(root) # keep adding root in stack.
+                root = root.left # traverse LEFT nodes at the end
+            
+            root = stack.pop()
+            res.append(root.val) # add ROOT to result.
+
+            root = root.right # traverse RIGHT nodes at the end
+
+        return res