# LeetCode 872. Leaf-Similar Trees

## Description

https://leetcode.com/problems/leaf-similar-trees/

Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence.

For example, in the given tree above, the leaf value sequence is `(6, 7, 4, 9, 8)`.

Two binary trees are considered leaf-similar if their leaf value sequence is the same.

Return `true` if and only if the two given trees with head nodes `root1` and `root2` are leaf-similar.

Example 1:

```Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8]
Output: true
```

Example 2:

```Input: root1 = , root2 = 
Output: true
```

Example 3:

```Input: root1 = , root2 = 
Output: false
```

Example 4:

```Input: root1 = [1,2], root2 = [2,2]
Output: true
```

Example 5:

```Input: root1 = [1,2,3], root2 = [1,3,2]
Output: false
```

Constraints:

• The number of nodes in each tree will be in the range `[1, 200]`.
• Both of the given trees will have values in the range `[0, 200]`.

## Explanation

Traverse both trees and collect leaf node values. Compare if two trees leaf node values are the same.

## Python Solution

``````# 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 leafSimilar(self, root1: TreeNode, root2: TreeNode) -> bool:

sequence1 = []
sequence2 = []

self.traverse(root1, sequence1)

self.traverse(root2, sequence2)

return sequence1 == sequence2

def traverse(self, root, results):
if not root:
return

if not root.left and not root.right:
results.append(root.val)
return

self.traverse(root.left, results)
self.traverse(root.right, results)``````
• Time Complexity: O(N).
• Space Complexity: O(N).