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

与问题非常类似，唯一区别在于这一次确定root的位置由post traversal来确定，为最后一个元素。

1 /** 2  * Definition for binary tree 3  * public class TreeNode { 4  *     int val; 5  *     TreeNode left; 6  *     TreeNode right; 7  *     TreeNode(int x) { val = x; } 8  * } 9  */10 public class Solution {11     public TreeNode buildTree(int[] inorder, int[] postorder) {12         if (inorder.length==0 || postorder.length==0 || inorder.length!=postorder.length) {13             return null;14         }15         int len = inorder.length;16         HashMap
     
       map = new HashMap
      
       ();17         for (int i=0; i
       
         map) {24         if (posL>posR || inL>inR) {25             return null;26         }27         TreeNode root = new TreeNode(postorder[posR]);28         int index = map.get(postorder[posR]);29         root.left = helper(postorder, posL, posL+index-1-inL, inorder, inL, index-1, map);30         root.right = helper(postorder, posL+index-inL, posR-1, inorder, index+1, inR, map);31         return root;32     }33 }