<p>Given two integer arrays <code>inorder</code> and <code>postorder</code> where <code>inorder</code> is the inorder traversal of a binary tree and <code>postorder</code> is the postorder 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> inorder = [9,3,15,20,7], postorder = [9,15,7,20,3]
<strong>Output:</strong> [3,9,20,null,null,15,7]
</pre>

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

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

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

<ul>
	<li><code>1 &lt;= inorder.length &lt;= 3000</code></li>
	<li><code>postorder.length == inorder.length</code></li>
	<li><code>-3000 &lt;= inorder[i], postorder[i] &lt;= 3000</code></li>
	<li><code>inorder</code> and <code>postorder</code> consist of <strong>unique</strong> values.</li>
	<li>Each value of <code>postorder</code> also appears in <code>inorder</code>.</li>
	<li><code>inorder</code> is <strong>guaranteed</strong> to be the inorder traversal of the tree.</li>
	<li><code>postorder</code> is <strong>guaranteed</strong> to be the postorder 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) {}
 * };
 */

TreeNode* makeTree(vector<vector<pair<int, bool>>> &adj, int u){
    TreeNode* node = new TreeNode(u);
    for(auto &&X: adj[u + 3000]){
        if(!X.second) node->left = makeTree(adj, X.first);
        else node->right = makeTree(adj, X.first);
    }
    return node;
}

class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int n = postorder.size();
        reverse(postorder.begin(), postorder.end());
        pair<int, bool> parent[6010];
        pair<int, int> lim[6010];
        int ind[6010];
        for(int i = 0; i < n; i++) ind[inorder[i] + 3000] = i;
        int index = ind[postorder[0] + 3000];
        lim[postorder[0] + 3000] = {0, n};
        for(int j = 0; j < index; j++) parent[inorder[j] + 3000] = {postorder[0], 0};
        for(int j = index + 1; j < n; j++) parent[inorder[j] + 3000] = {postorder[0], 1};
        for(int i = 1; i < n; i++){
            index = ind[postorder[i] + 3000];
            int parIndex = ind[parent[postorder[i] + 3000].first + 3000];
            int l = lim[parent[postorder[i] + 3000].first + 3000].first, u = lim[parent[postorder[i] + 3000].first + 3000].second;
            if(index > parIndex) l = parIndex + 1;
            else u = parIndex;
            lim[postorder[i] + 3000] = {l, u};
            for(int j = l; j < index; j++) parent[inorder[j] + 3000] = {postorder[i], 0};
            for(int j = index + 1; j < u; j++) parent[inorder[j] + 3000] = {postorder[i], 1};
        }
        vector<vector<pair<int, bool>>> adj(6010);
        for(int i = 1; i < n; i++) adj[parent[postorder[i] + 3000].first + 3000].push_back({postorder[i], parent[postorder[i] + 3000].second});
        TreeNode* root = makeTree(adj, postorder[0]);
        return root;
    }
};
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