<p>Consider all the leaves of a binary tree, from&nbsp;left to right order, the values of those&nbsp;leaves form a <strong>leaf value sequence</strong><em>.</em></p>

<p><img alt="" src="https://s3-lc-upload.s3.amazonaws.com/uploads/2018/07/16/tree.png" style="width: 400px; height: 336px;" /></p>

<p>For example, in the given tree above, the leaf value sequence is <code>(6, 7, 4, 9, 8)</code>.</p>

<p>Two binary trees are considered <em>leaf-similar</em>&nbsp;if their leaf value sequence is the same.</p>

<p>Return <code>true</code> if and only if the two given trees with head nodes <code>root1</code> and <code>root2</code> are leaf-similar.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/03/leaf-similar-1.jpg" style="width: 600px; height: 237px;" />
<pre>
<strong>Input:</strong> 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]
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/09/03/leaf-similar-2.jpg" style="width: 300px; height: 110px;" />
<pre>
<strong>Input:</strong> root1 = [1,2,3], root2 = [1,3,2]
<strong>Output:</strong> false
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in each tree will be in the range <code>[1, 200]</code>.</li>
	<li>Both of the given trees will have values in the range <code>[0, 200]</code>.</li>
</ul>

class Solution {
public:
    vector<int>v1,v2;
    int leafCount(TreeNode* p){
        int x,y;
        if(p){
            x = leafCount(p->left);
            y = leafCount(p->right);
            if(p->left==nullptr && p->right ==nullptr){
                 v1.push_back(p->val);
                 return x+y+1;
            }
            else{
                return x+y;
            }
        }
        return 0;
    }
        int leafCount1(TreeNode* p){
        int x,y;
        if(p){
            x = leafCount1(p->left);
            y = leafCount1(p->right);
            if(p->left==nullptr && p->right ==nullptr){
                 v2.push_back(p->val);
                 return x+y+1;
            }
            else{
                return x+y;
            }
        }
        return 0;
    }
    bool leafSimilar(TreeNode* root1, TreeNode* root2) {
          leafCount(root1);
          leafCount1(root2);
          if(v1==v2){
             return true;
          }
          return false;
    }
};
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