106. Construct Binary Tree from Inorder and Postorder Traversal（从中序与后序遍历序列构造二叉树）

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

思路同上一篇LeetCode105

只不过后续遍历是从后往前来确定根节点的，然后在inorder中找到root节点对应的下标。

如果不太理解下标的对应关系，可以先用例子：

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

Note:
You may assume that duplicates do not exist in the tree.

For example, given

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

Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

当在postorder中找到最后一个3，它是root节点，在inorder中对应index是1，然后inorder就被分割成了两部分，

左侧部分产生左子树，右侧部分产生右子树。依次递归下去，难点在于index与下一个递归边界的判断。

AC 8ms Java:

/**
 * 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) {
        return helper(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
    }
    public TreeNode helper(int[] inorder,int inStart,int inEnd,
                           int[] postorder,int poStart,int poEnd){
        if(inStart>inEnd||poStart>poEnd)
            return null;
        TreeNode root=new TreeNode(postorder[poEnd]);
        int index=0;
        for(int i=inStart;i<=inEnd;i++){
            if(inorder[i]==root.val){
                index=i;
                break;
            }
        }
        root.left=helper(inorder,inStart,index-1,postorder,poStart,poStart+index-inStart-1);
        root.right=helper(inorder,index+1,inEnd,postorder,poEnd-(inEnd-index),poEnd-1);
        return root;
    }
}