105. Construct Binary Tree from Preorder and Inorder Traversal
1. Description
Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.
2. Example
Example 1
Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]
Output: [3,9,20,null,null,15,7]
Example 2
Input: preorder = [-1], inorder = [-1]
Output: [-1]
3. Constraints
- 1 <= preorder.length <= 3000
- inorder.length == preorder.length
- -3000 <= preorder[i], inorder[i] <= 3000
- preorder and inorder consist of unique values.
- Each value of inorder also appears in preorder.
- preorder is guaranteed to be the preorder traversal of the tree.
- inorder is guaranteed to be the inorder traversal of the tree.
4. Solutions
Recursion
n = preorder.size()
Time complexity: O(n)
Space complexity: O(n)
class Solution {
public:
TreeNode *buildTree(const vector<int> &preorder, const vector<int> &inorder) {
const int n = preorder.size();
unordered_map<int, int> inorder_index;
for (int i = 0; i < n; ++i) {
inorder_index[inorder[i]] = i;
}
return build_tree(preorder, 0, n, 0, n, inorder_index);
}
private:
TreeNode *build_tree(
const vector<int> &preorder,
int pre_left,
int pre_right,
int in_left,
int in_right,
const unordered_map<int, int> &inorder_index) {
if (pre_left == pre_right) {
return nullptr;
} else {
TreeNode *root = new TreeNode(preorder[pre_left]);
int index = inorder_index.at(preorder[pre_left]);
root->left = build_tree(
preorder,
pre_left + 1,
pre_left + 1 + index - in_left,
in_left,
index,
inorder_index);
root->right = build_tree(
preorder,
pre_left + 1 + index - in_left,
pre_right,
index + 1,
in_right,
inorder_index);
return root;
}
}
};