題目：

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

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

思路：

中序遍歷的順序：左根右，後序遍歷的順序：左右根。因而我們可以找到最後一個結點是根節點，前序遍歷中找到此點，左邊是左子樹，右邊是右子樹，依次遞迴可以得到最終的結果。

程式碼：

/**
 * 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 || postorder==null || inorder.length==0|| postorder.length==0){
            return null
 
;
        }
        return constructTree(inorder,0,inorder.length-1,postorder,0,postorder.length-1);
    }
    public TreeNode constructTree(int[] inorder,int instart,int inend,int[] postorder,int poststart,int postend){
        if(instart>inend || poststart>postend){
            return null;
        }
        TreeNode root = new 
 
TreeNode(postorder[postend]);
        for(int i=0;i<postorder.length;i++){
            if(inorder[i] == postorder[postend]){
                root.left = constructTree(inorder,instart,i-1,postorder,poststart,poststart+i-1-instart);
                root.right = constructTree(inorder,i+1,inend,postorder,poststart+i-instart,postend-1);
            }
        }
        return root;
    }
}

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

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