leetcode 144.二叉树的前序遍历 Java

题目链接

https://leetcode-cn.com/problems/binary-tree-preorder-traversal/

描述

给定一个二叉树，返回它的 前序 遍历。
进阶: 递归算法很简单，你可以通过迭代算法完成吗？

示例

输入: [1,null,2,3]  
输出: [1,2,3]

初始代码模板

/**
 * 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) {
        
    }
}

代码

递归解法相当简单，代码如下

递归

class Solution {
    List<Integer> ans = new ArrayList<>();
    public List<Integer> preorderTraversal(TreeNode root) {
        dfs(root);
        return ans;
    }

    public void dfs(TreeNode root) {
        if (root == null) {
            return;
        }

        ans.add(root.val);
        dfs(root.left);
        dfs(root.right);
    }
}

使用迭代需要自行维护一个栈，清楚前序遍历的顺序，把遍历的节点的右节点和左节点相继压栈，代码如下

迭代

/**
 * 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;
        }

        Deque<TreeNode> stack = new ArrayDeque<>();
        stack.push(root);

        while (!stack.isEmpty()) {
            TreeNode node = stack.pop();

            res.add(node.val);
            if (node.right != null) {
                stack.push(node.right);
            } 
            if (node.left != null) {
                stack.push(node.left);
            }
        }

        return res;
    }
}