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feature: leaf similar trees dfs recursive solution
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from typing import Optional | ||
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from trees.TreeNode import TreeNode | ||
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# O(n) time || O(h) space | ||
def leaf_similar(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool: | ||
def dfs(node: TreeNode): | ||
if not node: | ||
return [] | ||
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if not node.left and not node.right: | ||
return [node.val] | ||
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return dfs(node.left) + dfs(node.right) | ||
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return dfs(root1) == dfs(root2) |
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import unittest | ||
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from trees.LeafSimilarTrees import leaf_similar | ||
from trees.TreeNode import TreeNode | ||
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class MyTestCase(unittest.TestCase): | ||
def test_leaf_similar(self): | ||
root1 = TreeNode( | ||
val=3, | ||
left=TreeNode( | ||
val=5, | ||
left=TreeNode(val=6), | ||
right=TreeNode( | ||
val=2, | ||
left=TreeNode(val=7), | ||
right=TreeNode(val=4) | ||
) | ||
), | ||
right=TreeNode( | ||
val=1, | ||
left=TreeNode(val=9), | ||
right=TreeNode(val=8) | ||
) | ||
) | ||
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root2 = TreeNode( | ||
val=3, | ||
left=TreeNode( | ||
val=5, | ||
left=TreeNode(val=6), | ||
right=TreeNode(val=7) | ||
), | ||
right=TreeNode( | ||
val=1, | ||
left=TreeNode(val=4), | ||
right=TreeNode( | ||
val=2, | ||
left=TreeNode(val=9), | ||
right=TreeNode(val=8) | ||
) | ||
) | ||
) | ||
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self.assertEqual(True, leaf_similar(self, root1, root2)) | ||
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def test_leaf_similar_1(self): | ||
root1 = TreeNode( | ||
val=1, | ||
left=TreeNode(val=2), | ||
right=TreeNode(val=3) | ||
) | ||
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root2 = TreeNode( | ||
val=1, | ||
left=TreeNode(val=3), | ||
right=TreeNode(val=2) | ||
) | ||
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self.assertEqual(False, leaf_similar(self, root1, root2)) | ||
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if __name__ == '__main__': | ||
unittest.main() |