Problem
Given a binary tree, return the inorder traversal of its nodes’ values.
For example:
Given binary tree [1,null,2,3],
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
Solution: Recursion
Time complexity: O(n)
Space complexity: O(h)
C++
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// Author: Huahua // Running time: 4 ms class Solution { public: vector<int> inorderTraversal(TreeNode* root) { vector<int> ans; inorderTraversal(root, ans); return ans; } private: void inorderTraversal(TreeNode* root, vector<int>& ans) { if (root == nullptr) return; inorderTraversal(root->left, ans); ans.push_back(root->val); inorderTraversal(root->right, ans); } }; |
Solution 2: Iterative
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// Author: Huahua // Running time: 3 ms lass Solution { public: vector<int> inorderTraversal(TreeNode* root) { if (root == nullptr) return {}; vector<int> ans; stack<TreeNode*> s; TreeNode* curr = root; while (curr != nullptr || !s.empty()) { while (curr != nullptr) { s.push(curr); curr = curr->left; } curr = s.top(); s.pop(); ans.push_back(curr->val); curr = curr->right; } return ans; } }; |
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