114. Flatten Binary Tree to Linked List

最新推荐文章于 2024-08-25 16:11:27 发布

N.C_IPOC_BUPT

最新推荐文章于 2024-08-25 16:11:27 发布

阅读量169

点赞数

CC 4.0 BY-SA版权

分类专栏：温故知新 Tree 我的Leetcode专栏

本文链接：https://blog.youkuaiyun.com/IPOC_BUPT/article/details/88418298

Tree 同时被 3 个专栏收录

84 篇文章

订阅专栏

我的Leetcode专栏

77 篇文章

订阅专栏

温故知新

53 篇文章

订阅专栏

本文详细解析了二叉树扁平化的三种高效算法：递归法、迭代法（使用栈）及Morris遍历。通过具体代码实现，展示了如何将二叉树转换为链表，同时保持节点的相对顺序，且不使用额外的数据结构。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目描述

在这里插入图片描述

方法思路

Approach1:

class Solution{
    //Runtime: 6 ms, faster than 99.92%
    //Memory Usage: 39.2 MB, less than 5.09%
  private TreeNode next = null;
  public void flatten(TreeNode root) {
    if (root == null) return;
      
    flatten(root.right);
    flatten(root.left);
      
    root.right = next;
    root.left = null;
      
    next = root;
  }
}

Approach2:

class Solution{
    //it is DFS so u need a stack.    
    //Dont forget to set the left child to null, or u'll get TLE. (tricky!)
    //Runtime: 7 ms, faster than 75.79%
    //Memory Usage: 38.2 MB, less than 11.03%
    public void flatten(TreeNode root) {
        if (root == null) return;
        Stack<TreeNode> stk = new Stack<TreeNode>();
        stk.push(root);
        while (!stk.isEmpty()){
            TreeNode curr = stk.pop();
            if (curr.right!=null)  
                 stk.push(curr.right);
            if (curr.left!=null)  
                 stk.push(curr.left);
            if (!stk.isEmpty()) 
                 curr.right = stk.peek();//查看此堆栈顶部的对象，而不从堆栈中删除它。 
            curr.left = null;  // dont forget this!! 
        }
    }
}

Approach3:

class Solution{
    //Runtime: 6 ms, faster than 99.92%
    //Memory Usage: 38.1 MB, less than 15.45% 
    //This solution is based on recursion. We simply flatten left and right subtree         and paste each sublist to the right child of the root. (don't forget to set left       child to null)
    public void flatten(TreeNode root) {
        if (root == null) return;
        
        TreeNode left = root.left;
        TreeNode right = root.right;
        
        root.left = null;
        
        flatten(left);
        flatten(right);
        
        root.right = left;
        TreeNode cur = root;
        while (cur.right != null) cur = cur.right;
        cur.right = right;
    }
}