diff --git a/data_structures/binary_tree/binary_tree_top_view.py b/data_structures/binary_tree/binary_tree_top_view.py
new file mode 100644
index 000000000000..73f84f724e25
--- /dev/null
+++ b/data_structures/binary_tree/binary_tree_top_view.py
@@ -0,0 +1,78 @@
+# Created by susmith98
+
+from collections import deque
+
+"""
+Problem Description:
+Given a binary tree, print it's top view from left to right.
+
+Solution Logic:
+Imagine binary tree as x axis from top view and root node is at index 0.
+
+If parent node is at x-axis index idx, then
+-> left child is at x-axis index idx-1
+-> right child is at x-axis index idx + 1
+
+Multiple nodes might be present at x-axis index idx on different depth's of tree.
+To tackle this scenario,
+Consider the node at the lowest depth in tree as we want the top view.
+To consider the node at the lowest depth, BFS is used.
+"""
+
+
+def topViewDict(
+    root: int,
+    binary_tree: dict = {},
+    x_axis_dict: dict = {},
+):
+    queue = deque()
+    # x axis index of root is 0
+    queue.append((root, 0))
+    while len(queue) > 0:
+        (parent, x_axis_idx) = queue.popleft()
+        if x_axis_idx not in x_axis_dict:
+            x_axis_dict[x_axis_idx] = parent
+        # Check if current node has children
+        if parent in binary_tree:
+            # Check for left child
+            if binary_tree[parent][0] is not None:
+                queue.append((binary_tree[parent][0], x_axis_idx - 1))
+            # Check for right child
+            if binary_tree[parent][1] is not None:
+                queue.append((binary_tree[parent][1], x_axis_idx + 1))
+    return
+
+
+def getTopViewListFromDict(x_axis_dict: dict = {}) -> list:
+    if len(x_axis_dict) == 0:
+        return []
+    x_axis_list = []
+    for idx in sorted(x_axis_dict.keys()):
+        x_axis_list.append(x_axis_dict[idx])
+    return x_axis_list
+
+
+def topView(root: int, binary_tree: dict = {}) -> list:
+    """
+    >>> topView(1, { 1: [2,3], 2: [4,5], 3: [6,7], 7: [8,9]})
+    [4, 2, 1, 3, 7, 9]
+    >>> topView(5, { 1: [2,3], 2: [4,5], 3: [6,7], 7: [8,9]})
+    []
+    >>> topView(1, { 1: [2,3], 2: [4,5], 3: [6,7], 4: [10,11], 5: [12,13],7: [8,9]})
+    [10, 4, 2, 1, 3, 7, 9]
+    """
+    if root not in binary_tree.keys() or len(binary_tree) == 0:
+        return []
+    x_axis_dict = {}
+    topViewDict(root, binary_tree, x_axis_dict=x_axis_dict)
+    return getTopViewListFromDict(x_axis_dict)
+
+
+if __name__ == "__main__":
+    binary_tree = {1: [2, 3], 2: [4, 5], 3: [6, 7], 4: [10, 11], 5: [12, 13], 7: [8, 9]}
+    root = 1
+    x_axis_list = topView(root, binary_tree)
+    print("Binary tree:", sep=" ")
+    print(binary_tree)
+    print("Top view from left to right:", sep=" ")
+    print(x_axis_list)