LeetCode OJ：Construct Binary Tree from Inorder and Postorder Traversal（从中序以及后序遍历结果中构造二叉树）

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

Note:
You may assume that duplicates do not exist in the tree.
同样是给出了不会有重复数字的条件，用递归较容易实现，代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        if(inorder.size() == 0)
            return NULL;
        return createTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);   
    }

    TreeNode* createTree(vector<int> & inorder, int inBegin, int inEnd,
     vector<int> & postorder, int postBegin, int postEnd)
    {
        if(inBegin > inEnd) return NULL;
        int rootVal = postorder[postEnd];
        int mid;
        for(int i = inBegin; i <= inEnd; ++i){
            if(inorder[i] == rootVal){
                mid = i;
                break;
            }
        }
        int len = mid - inBegin;
        TreeNode * left = createTree(inorder, inBegin, mid - 1,     //边界条件同样应该注意
                            postorder, postBegin, postBegin + len - 1);
        TreeNode * right = createTree(inorder, mid + 1, inEnd,
                            postorder, postBegin + len, postEnd - 1);
        TreeNode * root = new TreeNode(rootVal);
        root->left = left;
        root->right = right;
        return root;
    }
};

 java版本的代码如下所示，没有区别：

public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder.length == 0)
            return null;
        return createTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);       
    }
    
    public TreeNode createTree(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd){
        if(inEnd < inBegin)
            return null;
        int rootVal = postorder[postEnd];
        int mid = 0;
        for(int i = inBegin; i <= inEnd; ++i){
            if(inorder[i] == rootVal){
                mid = i;
                break;   
            }
        }
        int len = mid - inBegin;
        TreeNode leftNode = createTree(inorder, inBegin, mid - 1, postorder, postBegin, postBegin + len - 1);
        TreeNode rightNode = createTree(inorder, mid + 1, inEnd, postorder, postBegin + len, postEnd - 1);
        TreeNode root = new TreeNode(rootVal);
        root.left = leftNode;
        root.right = rightNode;
        return root;        
    }
}