105. 从前序与中序遍历序列构造二叉树

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如，给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]

返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

/**
 * 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>& preorder, vector<int>& inorder) 
    {
        int n = preorder.size();
        return buildTree(preorder, 0, n-1, inorder, 0, n-1);   
    }
    TreeNode* buildTree(vector<int> &preorder, int pleft, int pright, vector<int> &inorder, int ileft, int iright)
    {
        if(pleft > pright || ileft > iright)
            return NULL;
        TreeNode *curr = new TreeNode(preorder[pleft]);
        int i;
        for(i = ileft; i <= iright; ++i)
        {
            if(inorder[i] == curr->val)
                break;
        }
        curr->left = buildTree(preorder, pleft + 1, pleft + i -ileft, inorder, ileft, i -1);
        curr->right = buildTree(preorder, pleft + i - ileft + 1, pright, inorder, i + 1, iright);
        return curr;
    }
};