<p>Given two integer arrays, <code>preorder</code> and <code>postorder</code> where <code>preorder</code> is the preorder traversal of a binary tree of <strong>distinct</strong> values and <code>postorder</code> is the postorder traversal of the same tree, reconstruct and return <em>the binary tree</em>.</p>

<p>If there exist multiple answers, you can <strong>return any</strong> of them.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/24/lc-prepost.jpg" style="width: 304px; height: 265px;" />
<pre>
<strong>Input:</strong> preorder = [1,2,4,5,3,6,7], postorder = [4,5,2,6,7,3,1]
<strong>Output:</strong> [1,2,3,4,5,6,7]
</pre>

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

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

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

<ul>
	<li><code>1 &lt;= preorder.length &lt;= 30</code></li>
	<li><code>1 &lt;= preorder[i] &lt;= preorder.length</code></li>
	<li>All the values of <code>preorder</code> are <strong>unique</strong>.</li>
	<li><code>postorder.length == preorder.length</code></li>
	<li><code>1 &lt;= postorder[i] &lt;= postorder.length</code></li>
	<li>All the values of <code>postorder</code> are <strong>unique</strong>.</li>
	<li>It is guaranteed that <code>preorder</code> and <code>postorder</code> are the preorder traversal and postorder traversal of the same binary tree.</li>
</ul>

class Solution {
public:
    TreeNode*build(vector<int>&pre,int preLow,int preHigh,vector<int>&post,int postLow,int postHigh){
        // base case
        if(preLow > preHigh || postLow > postHigh) return NULL; 
        TreeNode*root=new TreeNode(pre[preLow]);
        // when there is only one node
        if(preLow == preHigh) return root;
        int i=postLow;
        // find the 2nd element of preOrder in postOrder 
        while(i <= postHigh){
            if(post[i]==pre[preLow+1]) break;
            i++;
        }
        int leftCount = i-postLow;
        // recursion part of left & right subtree
        root->left = build(pre, preLow+1, preLow+leftCount+1, post, postLow, i);
        root->right = build(pre, preLow+leftCount+2, preHigh, post, i+1, postHigh-1);
        return root;
    }
    TreeNode* constructFromPrePost(vector<int>& preorder, vector<int>& postorder) {
        int n=preorder.size();
        return build(preorder,0,n-1,postorder,0,n-1);
    }
};
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