144. Binary Tree Preorder Traversal

144. Binary Tree Preorder Traversal (Medium)

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?

Related Topics:
Stack, Tree

Similar Questions:

• Binary Tree Inorder Traversal (Medium)
• Verify Preorder Sequence in Binary Search Tree (Medium)
• N-ary Tree Preorder Traversal (Easy)

Solution 1. Recursive

// OJ: https://leetcode.com/problems/binary-tree-preorder-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
private:
    vector<int> v;
    void rec(TreeNode *root) {
        if (!root) return;
        v.push_back(root->val);
        rec(root->left);
        rec(root->right);
    }
public:
    vector<int> preorderTraversal(TreeNode* root) {
        rec(root);
        return v;
    }
};

Solution 2. Iterative

// OJ: https://leetcode.com/problems/binary-tree-preorder-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        if (!root) return {};
        vector<int> ans;
        stack<TreeNode*> s;
        s.push(root);
        while (s.size()) {
            root = s.top();
            s.pop();
            ans.push_back(root->val);
            if (root->right) s.push(root->right);
            if (root->left) s.push(root->left);
        }
        return ans;
    }
};

Solution 3. Iterative

// OJ: https://leetcode.com/problems/binary-tree-preorder-traversal/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    vector<int> preorderTraversal(TreeNode* root) {
        vector<int> ans;
        stack<TreeNode*> s;
        while (root || s.size()) {
            if (!root) {
                root = s.top();
                s.pop();
            }
            ans.push_back(root->val);
            if (root->right) s.push(root->right);
            root = root->left;
        }
        return ans;
    }
};