[LeetCode] 106. Construct Binary Tree from Inorder and Postorder Traversal

Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.

Thought process:
Similar to 105. Construct Binary Tree from Preorder and Inorder Traversal. Iterate from end to beginning of the post-order traversal. Attach right sub-tree to root before left.

Solution:
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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
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 buildTree(map, postorder, postorder.length - 1, 0, postorder.length - 1);
    }
    
    private TreeNode buildTree(Map<Integer, Integer> inorder, int[] postorder, int postIndex, int left, int right) {
        if (postIndex < 0 || left > right) {
            return null;
        }
        
        int inIndex = inorder.get(postorder[postIndex]);
        TreeNode node = new TreeNode(postorder[postIndex]);
        node.right = buildTree(inorder, postorder, postIndex - 1, inIndex + 1, right);
        node.left = buildTree(inorder, postorder, postIndex - (right -  inIndex + 1), left, inIndex - 1);
        return node;
    }
}

Time complexity: O(n).
Location: Philadelphia, PA, USA

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