Medium
Construct Binary Tree from Preorder and Inorder Traversal — C++
Full explanation · Time O(n) · Space O(n)
// Time: O(n)
// Space: O(n)
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
unordered_map<int, size_t> in_entry_idx_map;
for (size_t i = 0; i < inorder.size(); ++i) {
in_entry_idx_map.emplace(inorder[i], i);
}
return ReconstructPreInOrdersHelper(preorder, 0, preorder.size(), inorder, 0, inorder.size(),
in_entry_idx_map);
}
// Reconstructs the binary tree from pre[pre_s : pre_e - 1] and
// in[in_s : in_e - 1].
TreeNode *ReconstructPreInOrdersHelper(const vector<int>& preorder, size_t pre_s, size_t pre_e,
const vector<int>& inorder, size_t in_s, size_t in_e,
const unordered_map<int, size_t>& in_entry_idx_map) {
if (pre_s == pre_e || in_s == in_e) {
return nullptr;
}
auto idx = in_entry_idx_map.at(preorder[pre_s]);
auto left_tree_size = idx - in_s;
auto node = new TreeNode(preorder[pre_s]);
node->left = ReconstructPreInOrdersHelper(preorder, pre_s + 1, pre_s + 1 + left_tree_size,
inorder, in_s, idx, in_entry_idx_map);
node->right = ReconstructPreInOrdersHelper(preorder, pre_s + 1 + left_tree_size, pre_e,
inorder, idx + 1, in_e, in_entry_idx_map);
return node;
}
};