144. Binary Tree Preorder Traversal

难度: Easy

题目描述

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]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [1,2,4,5,6,7,3,8,9]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

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?

解题思路

代码实现

解决方案

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 LinkedList<>();
        LinkedList<TreeNode> stk = new LinkedList<>();
        TreeNode cur = root;
        while (cur != null || !stk.isEmpty()) {

            while (cur != null) {
                res.add(cur.val);
                if (cur.right != null) {
                    stk.push(cur.right);
                }

                cur = cur.left;

            }

            if (!stk.isEmpty()) {
                cur = stk.pop();
            }

        }

        return res;
    }

}