LeetCode OJ：Construct Binary Tree from Preorder and Inorder Traversal（从前序以及中序遍历结果中构造二叉树）

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

Note:
You may assume that duplicates do not exist in the tree.
从前序以及中序的结果中构造二叉树，这里保证了不会有两个相同的数字，用递归构造就比较方便了：

 class Solution {
 public:
     TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
         if(!preorder.size()) return NULL;
         return createTree(preorder, , preorder.size() - , inorder, , inorder.size() - );//边界条件应该想清楚
     }

     TreeNode* createTree(vector<int>& preOrder, int preBegin, int preEnd, vector<int>& inOrder, int inBegin, int inEnd)
     {
         if(preBegin > preEnd) return NULL;
         int rootVal = preOrder[preBegin];
         int mid;
         for(int i = inBegin; i <= inEnd; ++i)
             if(inOrder[i] == rootVal){
                 mid = i;
                 break;
             }
         int len = mid - inBegin;    //左边区间的长度为mid - inBegin;
         TreeNode * left = createTree(preOrder, preBegin + , preBegin + len, inOrder, inBegin, mid - );
         TreeNode * right = createTree(preOrder, preBegin + len + , preEnd, inOrder, mid + , inEnd);
         TreeNode * root = new TreeNode(rootVal);
         root->left = left;
         root->right = right;
         return root;
     }
 };

java版本的代码如下所示，方法上与上面的没什么区别：

 public class Solution {
     public TreeNode buildTree(int[] preorder, int[] inorder) {
         return createTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
     }

     public TreeNode createTree(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd){
         if(preBegin > preEnd)
             return null;
         int rootVal = preorder[preBegin];
         int i = inBegin;
         for(; i <= inEnd; ++i){
             if(inorder[i] == rootVal)
                 break;
         }
         int leftLen = i - inBegin;
         TreeNode root = new TreeNode(rootVal);
         root.left = createTree(preorder, preBegin + 1, preBegin + leftLen ,inorder, inBegin, i - 1); //这里的边界条件应该注意
         root.right = createTree(preorder, preBegin + 1 + leftLen, preEnd, inorder, i + 1, inEnd);
         return root;
     }
 }