leetcode-105-Construct Binary Tree from Preorder and Inorder Traversal

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

Note:

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

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

class Solution {
private:
    TreeNode* buildTree(const vector<int>& preorder, int p_l, int p_h, const vector<int>& inorder, int i_l, int i_h)
    {
        if(i_l > i_h) return NULL;
        int val = preorder[p_l];
        TreeNode *root = new TreeNode(val);
        int i = i_l;
        for(; i <= i_h; i++)
            if(inorder[i] == val) break;
        root->left = buildTree(preorder, p_l+1, p_l+i-i_l, inorder, i_l, i-1);
        root->right = buildTree(preorder, p_l+i-i_l+1, p_h, inorder, i+1, i_h);
        return root;
    }
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return buildTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
    }
};