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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 indices28 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 number43 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 }
Construct a tree from Inorder and Level order traversals