144

144. Binary Tree Preorder Traversal (Easy)

Given the root of a binary tree, return the preorder traversal of its nodes' values.

 

Example 1:

Input: root = [1,null,2,3]
Output: [1,2,3]

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?

Companies: Google, Apple, Microsoft

Related Topics:
Stack, Tree, Depth-First Search, Binary Tree

Similar Questions:

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

Solution 1. Recursive

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

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{{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;
    }
};