Construct Binary Tree from Inorder and Postorder Traversal
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0106.Construct%20Binary%20Tree%20from%20Inorder%20and%20Postorder%20Traversal/images/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>
Tree; Array; Hash Table; Divide and Conquer; Binary Tree
// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::cell::RefCell;
use std::collections::HashMap;
use std::rc::Rc;
impl Solution {
    pub fn build_tree(inorder: Vec<i32>, postorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
        let n = inorder.len();
        let mut d: HashMap<i32, usize> = HashMap::new();
        for i in 0..n {
            d.insert(inorder[i], i);
        }
        fn dfs(
            postorder: &[i32],
            d: &HashMap<i32, usize>,
            i: usize,
            j: usize,
            n: usize,
        ) -> Option<Rc<RefCell<TreeNode>>> {
            if n <= 0 {
                return None;
            }
            let val = postorder[j + n - 1];
            let k = *d.get(&val).unwrap();
            let left = dfs(postorder, d, i, j, k - i);
            let right = dfs(postorder, d, k + 1, j + k - i, n - 1 - (k - i));
            Some(Rc::new(RefCell::new(TreeNode { val, left, right })))
        }
        dfs(&postorder, &d, 0, 0, n)
    }
}
