<p>Given two binary search trees <code>root1</code> and <code>root2</code>, return <em>a list containing all the integers from both trees sorted in <strong>ascending</strong> order</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/12/18/q2-e1.png" style="width: 457px; height: 207px;" />
<pre>
<strong>Input:</strong> root1 = [2,1,4], root2 = [1,0,3]
<strong>Output:</strong> [0,1,1,2,3,4]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/12/18/q2-e5-.png" style="width: 352px; height: 197px;" />
<pre>
<strong>Input:</strong> root1 = [1,null,8], root2 = [8,1]
<strong>Output:</strong> [1,1,8,8]
</pre>

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

<ul>
	<li>The number of nodes in each tree is in the range <code>[0, 5000]</code>.</li>
	<li><code>-10<sup>5</sup> &lt;= Node.val &lt;= 10<sup>5</sup></code></li>
</ul>

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> getAllElements(TreeNode* root1, TreeNode* root2) {
        stack<TreeNode*> st1,st2;
        // st1.push(root1);st2.push(root2);
        vector<int> ans;
        while(root1 ||!st1.empty()|| root2 ||!st2.empty()){
            while(root1){
                st1.push(root1);
                root1=root1->left;
            }
            while(root2){
                st2.push(root2);
                root2=root2->left;
            }

            // focus on any one tree and hold the other traversal
            if(st2.empty() || (!st1.empty() && st1.top()->val<=st2.top()->val)){
                root1=st1.top();
                ans.push_back(root1->val);
                st1.pop();
                root1=root1->right;
            }
            // else(st1.empty()||(!st2.empty() && st2.top()->val<=st1.top()->val)){
            else{
                root2=st2.top();
                ans.push_back(root2->val);
                st2.pop();
                root2=root2->right;
            }


        }return ans;
    }
};
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