Description

Submission

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
    map<int, int> inorderIndex;

    TreeNode* dfs(int i, int j, int x, int y, vector<int>& preorder, vector<int>& inorder) {
        if(j - i < 0) return nullptr;
        if(j == i) return new TreeNode(preorder[i]);
        int rootIndexInorder = inorderIndex[preorder[i]];
        int lenLeft = rootIndexInorder - x;
        int lenRight = y - rootIndexInorder;
        return new TreeNode(
            preorder[i],
            dfs(i + 1, i + lenLeft, x, x + lenLeft - 1, preorder, inorder),
            dfs(i + lenLeft + 1, i + lenLeft + lenRight, rootIndexInorder + 1, rootIndexInorder + lenRight, preorder, inorder)
        );
    }
public:
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        for(int i = 0; i < inorder.size(); ++i) {
            inorderIndex[inorder[i]] = i;
        }

        int n = preorder.size();
        return dfs(0, n-1, 0, n-1, preorder, inorder);
    }
};

// range in preorder
// [i X X X X X j] X X X X

// r 
// 3,[9],[20,15,7]. i, j
//     r(rootIndexInorder)
// [9],3,[15,20,7] x y