<p>Given two integer arrays <code>preorder</code> and <code>inorder</code> where <code>preorder</code> is the preorder traversal of a binary tree and <code>inorder</code> is the inorder traversal of the same tree, construct and return <em>the binary tree</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/tree.jpg" style="width: 277px; height: 302px;" />
<pre>
<strong>Input:</strong> preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
<strong>Output:</strong> [3,9,20,null,null,15,7]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> preorder = [-1], inorder = [-1]
<strong>Output:</strong> [-1]
</pre>

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

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 3000</code></li>
	<li><code>inorder.length == preorder.length</code></li>
	<li><code>-3000 &lt;= preorder[i], inorder[i] &lt;= 3000</code></li>
	<li><code>preorder</code> and <code>inorder</code> consist of <strong>unique</strong> values.</li>
	<li>Each value of <code>inorder</code> also appears in <code>preorder</code>.</li>
	<li><code>preorder</code> is <strong>guaranteed</strong> to be the preorder traversal of the tree.</li>
	<li><code>inorder</code> is <strong>guaranteed</strong> to be the inorder traversal of the tree.</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:
    // Different Binary Trees.

    // Challenges:
    //.  Given a number/node, how to decide if this is left child, or right
    // child, or get back to parent. .  The answer is basicly coming from
    // inorder.

    // Use preorder to create child nodes.
    // Use inorder to decide if this is left or right child.
    // The main loop is around preorder.

    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        TreeNode* root = nullptr;
        std::stack<TreeNode*> parents;
        int preorder_index = 0;
        root = buildSubTree(preorder, inorder, preorder_index, 0, inorder.size()-1, parents, true);
        return root;
    }

    TreeNode* buildSubTree(vector<int>& preorder, vector<int>& inorder,
                           int& preorder_index, int inorder_x, int inorder_y,
                           std::stack<TreeNode*> parents, bool is_left) {

        if(inorder_x > inorder_y) {
            return nullptr;
        }

        TreeNode* new_node = new TreeNode(preorder[preorder_index]);

        if (parents.size() == 0) {
            parents.push(new_node); 
        } else if (is_left) {
            parents.top()->left = new_node;
            parents.push(new_node); 
        } else {
            parents.top()->right = new_node;
            parents.push(new_node); 
        }

        int target_index = inorder_x;
        while(inorder[target_index] != preorder[preorder_index] && target_index <= inorder_y) {
            target_index++;
        }
        preorder_index++;

        // left tree
        buildSubTree(preorder, inorder, preorder_index, inorder_x, target_index-1, parents, true);
        // right tree
        buildSubTree(preorder, inorder, preorder_index, target_index+1, inorder_y, parents,false);
        parents.pop();
        return new_node;
        
    }
};
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