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 = [1], root2 = [1]
Output: true
```

Example 3:

```Input: root1 = [1], root2 = [2]
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]`.

## Solution

Perform Depth-First Search on a tree and create an in-order list. Then, compare that list and the input list.

• Use a stack to keep track of the DFS path
• Check leaves 1 by 1, until the end or difference.

Time / space complexity: O(N)

```bool leafSimilar(TreeNode* root1, TreeNode* root2) {
stack<TreeNode*> s1 , s2;
s1.push(root1); s2.push(root2);
while (!s1.empty() && !s2.empty())
if (dfs(s1) != dfs(s2)) return false;
return s1.empty() && s2.empty();
}

int dfs(stack<TreeNode*>& s) {
while (true) {
TreeNode* node = s.top(); s.pop();
if (node->right) s.push(node->right);
if (node->left) s.push(node->left);
if (!node->left && !node->right) return node->val;
}
}```
```public boolean leafSimilar(TreeNode root1, TreeNode root2) {
Stack<TreeNode> s1 = new Stack<>(), s2 = new Stack<>();
s1.push(root1); s2.push(root2);
while (!s1.empty() && !s2.empty())
if (dfs(s1) != dfs(s2)) return false;
return s1.empty() && s2.empty();
}

public int dfs(Stack<TreeNode> s) {
while (true) {
TreeNode node = s.pop();
if (node.right != null) s.push(node.right);
if (node.left != null) s.push(node.left);
if (node.left == null && node.right == null) return node.val;
}
}```
```def leafSimilar(self, root1, root2):
def dfs(node):
if not node: return
if not node.left and not node.right: yield node.val
for i in dfs(node.left): yield i
for i in dfs(node.right): yield i
return all(a == b for a, b in itertools.izip_longest(dfs(root1), dfs(root2)))```
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