[Leetcode] Construct Binary Tree from Preorder and Inorder Traversal
2014-10-22 10:04
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题目:
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
思路:in order遍历后的数组为[left_tree, root, right_tree], preorder遍历后的数组为[root, left_tree, right_tree],由此可知,preorder数组的第一个元素必然为root,然后只需要查找到in order数组中对应node的位置,即可以将数组分割为属于左子树的node和属于右子树的node两部分。然后递归做即可。
总结:复杂度为O(n). 可以事先把inorder数组的元素以及index存到map中,这样以后的查找都可以在O(1)内完成。
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
思路:in order遍历后的数组为[left_tree, root, right_tree], preorder遍历后的数组为[root, left_tree, right_tree],由此可知,preorder数组的第一个元素必然为root,然后只需要查找到in order数组中对应node的位置,即可以将数组分割为属于左子树的node和属于右子树的node两部分。然后递归做即可。
class Solution { public: unordered_map<int, int> nodes; TreeNode* build_tree_helper(const vector<int>& preorder, const vector<int>& inorder, int pre_begin, int pre_end, int in_begin, int in_end) { if (pre_begin < 0 || in_begin < 0 || pre_end >= preorder.size() || in_end >= inorder.size() || pre_begin > pre_end || in_begin > in_end) return nullptr; TreeNode* new_root = new TreeNode(preorder[pre_begin]); int index = nodes.find(preorder[pre_begin])->second; new_root->left = build_tree_helper(preorder, inorder, pre_begin + 1, pre_begin + index - in_begin, in_begin, index - 1); new_root->right = build_tree_helper(preorder, inorder, pre_begin + index - in_begin + 1, pre_end, index + 1, in_end); return new_root; } TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) { for (int i = 0; i < inorder.size(); ++i) { nodes.insert(make_pair(inorder[i], i)); } return build_tree_helper(preorder, inorder, 0, preorder.size() - 1, 0, inorder.size() - 1); } };
总结:复杂度为O(n). 可以事先把inorder数组的元素以及index存到map中,这样以后的查找都可以在O(1)内完成。
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