106. Construct Binary Tree from Inorder and Postorder Traversal

```java
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }

inorder =   [9,3,15,20,7], 
postorder = [9,15,7,20,3]

postorder last one is root

find 3 in inorder
left is 9, right is [15, 20, 7]


build a hashmpa for inorder to find fast

T: O(n)
S: O(n)
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            map.put(inorder[i], i);
        }
        return dfs(inorder, postorder, 0, postorder.length-1, 0, inorder.length-1, map);
    }
    private TreeNode dfs(int[] inorder, int[] postorder, int postS, int postE, int inS, int inE, Map<Integer, Integer> map) {
        if (postS > postE) {
            return null;
        }
        int rootNumber = postorder[postE];
        int rootIndex = map.get(rootNumber);

        TreeNode root = new TreeNode(rootNumber);
        int leftSize = rootIndex - inS;
        root.left = dfs(inorder, postorder, postS, postS + leftSize - 1, inS, rootIndex-1, map); // post [left, right, root]
        root.right = dfs(inorder, postorder, postS + leftSize, postE - 1, rootIndex+1, inE, map);
        return root;
    }
}

```