Construct Binary Tree from Inorder and Postorder Traversal

https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

Given inorder and postorder traversal of a tree, construct the binary tree.

public TreeNode buildTree(int[] preorder, int[] inorder)

这一题和 Preorder Inorder建树的那题没有太多区别，代码相似度基本90%。原理还是那样，通过找根节点的位置在postorder和inorder相对的位置来确定左右子树在数组的范围，然后不停往下递归。唯一的不同点就在于，preorder的根节点永远在数组或者数组子集的第一个位置，而postorder的根节点永远在数组或者其子集的最后一位。其余情况相同，那么算法步骤就不仔细说了，详细请看Construct Binary Tree from Preorder and Inorder Traversal。

    public TreeNode buildTree(int[] inorder, int[] postorder) {        return helper(inorder, postorder, 0, inorder.length - 1, 0, postorder.length - 1);    }        public TreeNode helper(int[] inorder, int[] postorder, int in_start, int in_end, int post_start, int post_end){        if(post_start > post_end)            return null;        if(post_start == post_end)            return new TreeNode(postorder[post_start]);                    int in_pos = 0;        for(int i = in_start; i <= in_end; i++){            if(postorder[post_end] == inorder[i]){//这里是唯一的不同，preorder用pre_start找，而postorder用post_end来找                in_pos = i;                break;            }        }                TreeNode root = new TreeNode(postorder[post_end]);        root.left = helper(inorder, postorder, in_start, in_pos - 1, post_start, post_start + (in_pos - in_start - 1));        root.right = helper(inorder, postorder, in_pos + 1, in_end, post_start + (in_pos - in_start), post_end - 1);        return root;    }