Construct a tree from Inorder and Level order traversals

Given inorder and level-order traversals of a Binary Tree, construct the Binary Tree. Following is an example to illustrate the problem.

BinaryTree

Input: Two arrays that represent Inorder and level order traversals of a Binary Tree
in[]    = {4, 8, 10, 12, 14, 20, 22};
level[] = {20, 8, 22, 4, 12, 10, 14};

Output: Construct the tree represented by the two arrays. For the above two arrays, the constructed tree is shown in the diagram.

geeksforgeeks的做法是，每次以in和level数组去构建以level[0]为根结点的树。生成下一次level结点的开销是O(n)，所以整个时间复杂度是O(n^2)。

我的做法是：

1. 先计算出所有点的层序号。基于这个规律，如果两个元素在同一层，那么后面的数在中序遍历的顺序中，必然也是处于后面；如果后面的数在中序遍历中处于前面，那么必然是处于下一层。O(n)可以做到，但是需要先对两个数组作索引。

2. 从最后一层开始，每一层的左结点，是在inorder序列中，在它左边的连续序列（该序列必须保证层数比它大）中第一个层数=它的层数+1的数。右结点同理。查找左右结点的开销需要O(n)。

所以最终可以做到$O(n^2)$。

 1 struct TreeNode {
 2     int val;
 3     TreeNode *left, *right;
 4     TreeNode(int v): val(v), left(NULL), right(NULL) {}
 5 };
 6
 7 void print(TreeNode *root) {
 8     if (root == NULL) {
 9         cout << "NULL ";
10     } else {
11         cout << root->val << " ";
12         print(root->left);
13         print(root->right);
14     }
15 }
16
17 struct Indices {
18     int inOrderIndex;
19     int levelOrderIndex;
20     int level;
21 };
22
23 int main(int argc, char** argv) {
24     vector<int> inOrder = {4, 8, 10, 12, 14, 20, 22};
25     vector<int> levelOrder = {20, 8, 22, 4, 12, 10, 14};
26
27     // build indices
28     unordered_map<int, Indices> indices;
29     for (int i = 0; i < inOrder.size(); ++i) {
30         if (indices.count(inOrder[i]) <= 0) {
31             indices[inOrder[i]] = {i, 0, 0};
32         } else {
33             indices[inOrder[i]].inOrderIndex = i;
34         }
35         if (indices.count(levelOrder[i]) <= 0) {
36             indices[levelOrder[i]] = {0, i, 0};
37         } else {
38             indices[levelOrder[i]].levelOrderIndex = i;
39         }
40     }
41
42     // get level no. for each number
43     int level = 0;
44     for (int i = 1; i < levelOrder.size(); ++i) {
45         if (indices[levelOrder[i]].inOrderIndex < indices[levelOrder[i - 1]].inOrderIndex) {
46             ++level;
47         }
48         indices[levelOrder[i]].level = level;
49     }
50
51     unordered_map<int, TreeNode*> nodes;
52     for (int i = levelOrder.size() - 1; i >= 0; --i) {
53         nodes[levelOrder[i]] = new TreeNode(levelOrder[i]);
54         int index = indices[levelOrder[i]].inOrderIndex;
55         for (int j = index - 1; j >= 0 && indices[inOrder[j]].level > indices[inOrder[index]].level; --j) {
56             if (indices[inOrder[j]].level == indices[inOrder[index]].level + 1) {
57                 nodes[levelOrder[i]]->left = nodes[inOrder[j]];
58                 break;
59             }
60         }
61         for (int j = index + 1; j < levelOrder.size() && indices[inOrder[j]].level > indices[inOrder[index]].level; ++j) {
62             if (indices[inOrder[j]].level == indices[inOrder[index]].level + 1) {
63                 nodes[levelOrder[i]]->right = nodes[inOrder[j]];
64                 break;
65             }
66         }
67     }
68     print(nodes[levelOrder[0]]);
69     cout << endl;
70     return 0;
71 }