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[Easy - wk4 - Q1] 94. Binary Tree Inorder Traversal
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94. Binary Tree Inorder Traversal
Given the root of a binary tree, return the inorder traversal of its nodes' values.
Example 1:
Input: root = [1,null,2,3] Output: [1,3,2]
Example 2:
Input: root = [] Output: []
Example 3:
Input: root = [1] Output: [1]
Constraints:
- The number of nodes in the tree is in the range
[0, 100]. -100 <= Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
태그
#leetcode, #문제풀이, #easy, #microsoft, #amazon, #adobe, #apple, #yahoo, #stack, #tree, #depth-first search, #binary tree
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mingki님의 댓글
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C++
Runtime: 3 ms, faster than 56.18% of C++ online submissions for Binary Tree Inorder Traversal.
Memory Usage: 10.3 MB, less than 7.18% of C++ online submissions for Binary Tree Inorder Traversal.
Runtime: 3 ms, faster than 56.18% of C++ online submissions for Binary Tree Inorder Traversal.
Memory Usage: 10.3 MB, less than 7.18% of C++ online submissions for Binary Tree Inorder Traversal.
class Solution {
vector<int> v;
public:
vector<int> inorderTraversal(TreeNode* root) {
if (root) {
inorderTraversal(root->left);
v.push_back(root->val);
inorderTraversal(root->right);
}
return v;
}
};
