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.

class Solution {

public:
    /* from Preorder and Inorder Traversal */
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {

        return helper(preorder,0,preorder.size(),inorder,0,inorder.size());

    }

    TreeNode* helper(vector<int>& preorder,int i,int j,vector<int>& inorder,int ii,int jj)
    {
        // tree        8 4 5 3 7 3
        // preorder    8 [4 3 3 7] [5]
        // inorder     [3 3 4 7] 8 [5]

        // 每次从 preorder 头部取一个值 mid，作为树的根节点
        // 检查 mid 在 inorder 中 的位置，则 mid 前面部分将作为 树的左子树，右部分作为树的右子树

        if(i >= j || ii >= j)
            return NULL;

        int mid = preorder[i];
        auto f = find(inorder.begin() + ii,inorder.begin() + jj,mid);

        int dis = f - inorder.begin() - ii;

        TreeNode* root = new TreeNode(mid);
        root -> left = helper(preorder,i + 1,i + 1 + dis,inorder,ii,ii + dis);
        root -> right = helper(preorder,i + 1 + dis,j,inorder,ii + dis + 1,jj);
        return root;
    }
};