Construct Binary Tree from Preorder and Inorder Traversal

Q:

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

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

 

A:

/**
 * Definition for binary tree
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (inorder.empty() || preorder.empty()) return NULL;
        return buildTreeInternal(inorder, 0, inorder.size() - 1,
                                 preorder, 0);
    }
private:
    TreeNode *buildTreeInternal(vector<int> &inorder, int start, int end,
                        vector<int> &preorder, int root_pos_pre) {
        if (start > end) return NULL;
        int root_pos_in = -1;
        for (int i = start; i <= end; ++i) {
            if (inorder[i] == preorder[root_pos_pre]) {
                root_pos_in = i;
                break;
            }
        }
        if (-1 == root_pos_in) {
            assert(0);
            return NULL;
        }
        TreeNode* root = new TreeNode(preorder[root_pos_pre]);
        if (start <= root_pos_in - 1) {
            root->left = buildTreeInternal(inorder, start, root_pos_in - 1,
                                           preorder, root_pos_pre + 1);
            root_pos_pre += root_pos_in - start;
        }
        if (end >= root_pos_in + 1) {
            root->right = buildTreeInternal(inorder, root_pos_in + 1, end,
                                            preorder, root_pos_pre + 1);
            
        }

        return root;
    }
};