Leetcode: 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.

Example:

inorder: DBEFAGC

postorder: DFEBGCA

Root start with the last element of postorder array, A. In inorder array, all elements going before A are left child, and all elements following after A are right child. And the next right root start with C, left root start with B, both of whose position can be inferred from position of A in inorder and postorder array.

/** * Definition for binary tree * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public TreeNode buildTree(int[] inorder, int[] postorder) {        if (inorder == null || inorder.length == 0 || postorder == null || postorder.length == 0) {            return null;        }                return helperTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);    }        private TreeNode helperTree(int[] inorder, int instart, int inend, int[] postorder, int poststart, int postend) {        if (instart > inend) {            return null;        }                TreeNode root = new TreeNode(postorder[postend]);        int pos = findPos(inorder, instart, inend, postorder[postend]);        root.right = helperTree(inorder, pos + 1, inend, postorder, poststart + pos - instart, postend - 1);        root.left = helperTree(inorder, instart, pos - 1, postorder, poststart, poststart + pos - instart - 1);        return root;    }        private int findPos(int[] array, int start, int end, int key) {        for (int i = start; i <= end; i++) {            if (array[i] == key) {                return i;            }        }                return -1;    }}