LeetCode-144. Binary Tree Preorder Traversal [C++][Java]

LeetCode-144. Binary Tree Preorder TraversalLevel up your coding skills and quickly land a job. This is the best place to expand your knowledge and get prepared for your next interview.https://leetcode.com/problems/binary-tree-preorder-traversal/

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?

【C++】

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

【Java】

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<>();
        if (root == null) {return res;}
        Stack<TreeNode> s = new Stack<>();
        s.push(root);
        while (!s.isEmpty()) {
            TreeNode node = s.pop();
            res.add(node.val);
            if (node.right != null) {s.push(node.right);}
            if (node.left != null) {s.push(node.left);}
        }
        return res;
    }
}