2021-11-21 Daily-Challenge
Today I have done leetcode's November LeetCoding Challenge with cpp
.
November LeetCoding Challenge 21
Description
Construct Binary Tree from Inorder and Postorder Traversal
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.
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]
Constraints:
1 <= inorder.length <= 3000
postorder.length == inorder.length
-3000 <= inorder[i], postorder[i] <= 3000
inorder
andpostorder
consist of unique values.- Each value of
postorder
also appears ininorder
. inorder
is guaranteed to be the inorder traversal of the tree.postorder
is guaranteed to be the postorder traversal of the tree.
Solution
auto speedup = [](){
cin.tie(nullptr);
cout.tie(nullptr);
ios::sync_with_stdio(false);
return 0;
}();
class Solution {
TreeNode* buildTree(
vector<int> &inorder,
int inBegin,
int inEnd,
vector<int> &postorder,
int postBegin,
int postEnd
) {
if(inBegin > inEnd) return nullptr;
TreeNode *root = new TreeNode(postorder[postEnd]);
for(int i = inBegin; i <= inEnd; ++i) {
if(inorder[i] == postorder[postEnd]) {
int offset = i - inBegin;
root->left = buildTree(inorder, inBegin, inBegin + offset - 1, postorder, postBegin, postBegin + offset - 1);
root->right = buildTree(inorder, inBegin + offset + 1, inEnd, postorder, postBegin + offset, postEnd - 1);
break;
}
}
return root;
}
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
int len = inorder.size();
return buildTree(inorder, 0, len - 1, postorder, 0, len - 1);
}
};
// Accepted
// 202/202 cases passed (32 ms)
// Your runtime beats 32.56 % of cpp submissions
// Your memory usage beats 62.12 % of cpp submissions (26.2 MB)