[leetcode]94. Binary Tree Inorder Traversal@Java解题报告-优快云博客

本文链接：https://blog.youkuaiyun.com/zjkC050818/article/details/76665058

本文介绍了一种非递归方式实现二叉树中序遍历的方法，并提供了详细的Java实现代码。通过使用栈来辅助完成遍历过程，避免了递归带来的额外开销。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

https://leetcode.com/problems/binary-tree-inorder-traversal/description/

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

For example:
Given binary tree [1,null,2,3],

return [1,3,2].

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

题意是用非递归的方式实现中序遍历，代码很简单，如下：

package go.jacob.day804;

import java.util.ArrayList;
import java.util.List;
import java.util.Stack;

public class Demo1 {
	/*
	 * 非递归实现二叉树的中序遍历
	 */
	public List<Integer> inorderTraversal(TreeNode root) {
		Stack<TreeNode> stack = new Stack<TreeNode>();
		List<Integer> res = new ArrayList<Integer>();
		TreeNode node = root;
		while (!stack.isEmpty() || node != null) {
			while (node != null) {
				stack.push(node);
				node = node.left;
			}

			node = stack.pop();
			res.add(node.val);
			node = node.right;
			continue;

		}
		return res;
	}

	/*
	 * My Solution:二叉树中序遍历（循环实现）
	 */
	public List<Integer> inorderTraversal_byme(TreeNode root) {
		Stack<TreeNode> stack = new Stack<TreeNode>();
		List<Integer> res = new ArrayList<Integer>();
		TreeNode node = root;
		while (!stack.isEmpty() || node != null) {
			if (node == null) {
				node = stack.pop();
				res.add(node.val);
				node = node.right;
				continue;
			}
			stack.push(node);
			node = node.left;
		}
		return res;
	}

	private class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;

		TreeNode(int x) {
			val = x;
		}
	}
}