Given a binary tree, return the preorder traversal of its nodes’ values.
Example:
Input:[1,null,2,3]1 \ 2 / 3 Output:[1,2,3]
Follow up: Recursive solution is trivial, could you do it iteratively?
Solution 1: Recursion
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> ans; function<void(TreeNode*)> preorder = [&](TreeNode* n) { if (!n) return; ans.push_back(n->val); preorder(n->left); preorder(n->right); }; preorder(root); return ans; } }; |
Solution 2: Stack
Time complexity: O(n)
Space complexity: O(n)
C++
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// Author: Huahua class Solution { public: vector<int> preorderTraversal(TreeNode* root) { vector<int> ans; stack<TreeNode*> s; if (root) s.push(root); while (!s.empty()) { TreeNode* n = s.top(); ans.push_back(n->val); s.pop(); if (n->right) s.push(n->right); if (n->left) s.push(n->left); } return ans; } |
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