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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 }
Construct a tree from Inorder and Level order traversals
标签：des style blog http color ar sp for div
原文地址：http://www.cnblogs.com/linyx/p/4085784.html
