19.3.5 [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 {
public:
    int p1;
    TreeNode*buildchild(vector<int>& preorder, vector<int>& inorder, int i1, int i2) {
        if (i2 < i1)
            return NULL;
        int nowroot = preorder[p1];
        TreeNode*ans = new TreeNode(nowroot);
        for (int i = i1; i <= i2; i++)
            if (inorder[i] == nowroot) {
                if (i != i1) {
                    ++p1;
                    ans->left = buildchild(preorder, inorder, i1, i - 1);
                }
                if (i != i2) {
                    ++p1;
                    ans->right = buildchild(preorder, inorder, i + 1, i2);
                }
            }
        return ans;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        p1 = 0;
        return buildchild(preorder, inorder, 0, inorder.size() - 1);
    }
};