iteration Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/iteration/ Thu, 30 Jan 2020 01:34:53 +0000 en hourly 1 https://wordpress.org/?v=6.0.8 https://zxi.mytechroad.com/blog/wp-content/uploads/2017/09/cropped-photo-32x32.jpg iteration Archives - Huahua's Tech Road https://zxi.mytechroad.com/blog/tag/iteration/ 32 32 花花酱 LeetCode 144. Binary Tree Preorder Traversal https://zxi.mytechroad.com/blog/tree/leetcode-144-binary-tree-preorder-traversal/ https://zxi.mytechroad.com/blog/tree/leetcode-144-binary-tree-preorder-traversal/#respond Thu, 30 Jan 2020 01:34:46 +0000 https://zxi.mytechroad.com/blog/?p=6165 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…

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]]> 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++

// 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++

// 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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