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