<p>Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.</p>

<p>According to the <a href="https://en.wikipedia.org/wiki/Lowest_common_ancestor" target="_blank">definition of LCA on Wikipedia</a>: &ldquo;The lowest common ancestor is defined between two nodes <code>p</code> and <code>q</code> as the lowest node in <code>T</code> that has both <code>p</code> and <code>q</code> as descendants (where we allow <b>a node to be a descendant of itself</b>).&rdquo;</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong> The LCA of nodes 5 and 1 is 3.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;" />
<pre>
<strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
<strong>Output:</strong> 5
<strong>Explanation:</strong> The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [1,2], p = 1, q = 2
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 10<sup>5</sup>]</code>.</li>
	<li><code>-10<sup>9</sup> &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
	<li>All <code>Node.val</code> are <strong>unique</strong>.</li>
	<li><code>p != q</code></li>
	<li><code>p</code> and <code>q</code> will exist in the tree.</li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    bool path(TreeNode* root, int x, vector< TreeNode*>& arr){
        if(root==NULL) return false;
        arr.push_back(root);
        if(root->val==x){
            return true;
        }
        if(path(root->left,x,arr) || path(root->right,x,arr)) return true;
         arr.pop_back();
    return false; 
    }



public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        vector< TreeNode*> ans1;
        vector< TreeNode*> ans2;
        path(root,p->val,ans1);
        path(root,q->val,ans2);
        int flag=0;
        int i = ans1.size()-1;
        int j = ans2.size()-1;
        for(i=ans1.size()-1; i>=0; i--){
            for(j=ans2.size()-1; j>=0; j--){
                if(ans1[i]->val==ans2[j]->val){
                    flag=1;
                    break;
                }
            }
            if(flag==1) break;
        }
        return ans1[i];
    }
};
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