CODE 39: Submission Details

最新推荐文章于 2025-08-16 17:12:46 发布

原创最新推荐文章于 2025-08-16 17:12:46 发布 · 573 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 专栏收录该内容

154 篇文章

订阅专栏

本文介绍了一种不使用递归的方法来完成二叉树节点值的中序遍历。通过栈数据结构实现了迭代式的中序遍历算法，并提供了详细的Java实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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

For example:
Given binary tree {1,#,2,3},

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

	public ArrayList<Integer> inorderTraversal(TreeNode root) {
		// Start typing your Java solution below
		// DO NOT write main() function
		if (null == root) {
			return new ArrayList<Integer>();
		}
		ArrayList<Integer> inorder = new ArrayList<Integer>();
		Stack<TreeNode> stack = new Stack<TreeNode>();
		TreeNode node = root;
		while (!stack.isEmpty() || null != node) {
			if (null == node) {
				node = stack.pop();
				inorder.add(node.val);
				node = node.right;
			} else {
				stack.push(node);
				node = node.left;
			}
		}
		return inorder;
	}