【Leetcode】Binary Tree Inorder Traversal

最新推荐文章于 2025-08-14 09:11:10 发布

转载最新推荐文章于 2025-08-14 09:11:10 发布 · 173 阅读

文章标签：

#java #leetcode #ArrayList #binary tree

本文介绍了一种实现二叉树中序遍历的方法，包括使用递归和迭代两种方式，并提供了详细的Java代码实现。文章重点介绍了如何利用栈来完成非递归的中序遍历，以及Morris算法的实现。

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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}".

二叉树中序遍历

Java:

1. http://blog.youkuaiyun.com/fightforyourdream/article/details/16857347

/** 
 * Definition for binary tree 
 * public class TreeNode { 
 *     int val; 
 *     TreeNode left; 
 *     TreeNode right; 
 *     TreeNode(int x) { val = x; } 
 * } 
 */  
public class Solution {  
    public ArrayList<Integer> inorderTraversal(TreeNode root) {  
        ArrayList<Integer> ret = new ArrayList<Integer>();  
        if(root == null){  
            return ret;  
        }  
          
        Stack<TreeNode> stack = new Stack<TreeNode>();  
        TreeNode cur = root;  
        while(true){  
            while(cur != null){  
                stack.push(cur);  
                cur = cur.left;  
            }  
            if(stack.isEmpty()){  
                break;  
            }  
            cur = stack.pop();  
            ret.add(cur.val);  
            cur = cur.right;  
        }  
        return ret;  
    }  
}

2. 水印人生：

Solution 1: Use a stack. Time:O(n) and space O(logn).

</pre><pre name="code" class="java"><span style="font-family: Arial, Helvetica, sans-serif;">public class Solution{ </span><pre name="code" class="java">public  ArrayList<Integer> inorderTraversal(TreeNode root) {
            // Start typing your Java solution below
            // DO NOT write main() function
            ArrayList<Integer> res = new ArrayList<Integer>();
            if(root==null) return res;
            
            Stack<TreeNode> s = new Stack<TreeNode>();
            TreeNode cur = root;
            while(!s.isEmpty()||cur!=null){
                if(cur!=null){
                    s.push(cur); 
                    cur=cur.left;
                }else{
                    cur=s.pop();
                    res.add(cur.val);
                    cur=cur.right;
                }
            }
            return res;
        }
}

Solution 2: Morris algorithm. Time: O(nlogn) but space O(1).

public class Solution {
    public ArrayList<Integer> inorderTraversal(TreeNode root) {
        // Start typing your Java solution below
        // DO NOT write main() function
        ArrayList<Integer> res = new ArrayList<Integer>();
        TreeNode cur=root,next=null;
        while(cur!=null){
            if(cur.left!=null){
                next=cur;
                cur=cur.left;
            
                TreeNode temp=cur;
                while(temp.right!=null && temp.right!=next){
                    temp=temp.right;
                }    
            
                if(temp.right==null){
                    temp.right=next;
                }else{
                    temp.right=null;
                    res.add(next.val);
                    cur=next.right;
                }
            }else{
                res.add(cur.val);
                cur=cur.right;
            }
        }
        return res;
    }
}