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() == )
             return NULL;
         return createTree(inorder, , inorder.size() - , postorder, , postorder.size() - );
     }

     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 - ,     //边界条件同样应该注意
                             postorder, postBegin, postBegin + len - );
         TreeNode * right = createTree(inorder, mid + , inEnd,
                             postorder, postBegin + len, postEnd - );
         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;
     }
 }