144. Binary Tree Preorder Traversal

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144. Binary Tree Preorder Traversal

Description

Difficulty: Easy

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

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

Example 1:

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

Example 2:

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Input: root = []
Output: []

Example 3:

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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?

Hints/Notes

  • Use stack to solve it iteratively

Solution

Language: C++

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> res;

    vector<int> preorderTraversal(TreeNode* root) {
        if (root == nullptr) {
            return res;
        }
        res.push_back(root->val);
        preorderTraversal(root->left);
        preorderTraversal(root->right);
        return res;
    }
};

stack solution

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class Solution {
public:
    vector<int> res;

    vector<int> preorderTraversal(TreeNode* root) {
        stack<TreeNode*> s;
        if (root) s.push(root);
        while (!s.empty()) {
            TreeNode* node = s.top();
            s.pop();
            res.push_back(node->val);
            if (node->right) s.push(node->right);
            if (node->left) s.push(node->left);
        }
        return res;
    }
};