LeetCode题解：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.

题解：通过中序遍历和后序遍历还原二叉树

解决思路：首先要明确一点，对于后序遍历的结果，如果一个元素所在的位置为i，若在中序遍历的i-1位置的元素为该元素的根结点，说明该元素就是所在子树的右儿子（且没有子树），否则存在右子树。左子树倒没什么特别的

代码：

public class Solution {    private int inLen;    private int postLen;    public TreeNode buildTree(int[] inorder, int[] postorder) {        inLen = inorder.length;        postLen = postorder.length;        return buildTree(inorder, postorder, null);    }    private TreeNode buildTree(int[] inorder, int[] postorder, TreeNode root){        if(postLen < 0){            return null;        }        TreeNode node = new TreeNode(postorder[postLen--]);        if(inorder[postLen] != node.val){            node.right = buildTree(inorder, postorder, node);        }        inLen--;        if((root == null) || inorder[inLen] != root.val){            node.left = buildTree(inorder, postorder, node);        }        return node;    }}