【Leetcode】Binary Tree Postorder 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 · 166 阅读

文章标签：

#java #leetcode #binary tree #Tree node #ArrayList

本文介绍了一种实现二叉树后序遍历的方法，包括递归和迭代两种方式。递归方法简洁但可能引起栈溢出，迭代方法使用栈来避免递归带来的问题。

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

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

return [3,2,1].

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

Java:

Iteration: http://blog.youkuaiyun.com/linhuanmars/article/details/22009351

public List<Integer> postorderTraversal(TreeNode root) {
    List<Integer> res = new ArrayList<Integer>();
    if(root == null)
    {
        return res;
    }
    LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
    TreeNode pre = null;
    while(root != null || !stack.isEmpty())
    {
        if(root!=null)
        {
            stack.push(root);
            root = root.left;
        }
        else
        {
            TreeNode peekNode = stack.peek();
            if(peekNode.right!=null && pre != peekNode.right)
            {
                root = peekNode.right;
            }
            else
            {
                stack.pop();
                res.add(peekNode.val);
                pre = peekNode;
            }
        }
    }
    return res;
}

Recursion:

http://blog.youkuaiyun.com/xiaozhuaixifu/article/details/17101265

public class Solution {  
    public ArrayList<Integer> postorderTraversal(TreeNode root) {  
        ArrayList<Integer> res = new ArrayList<Integer>();  
         if(root == null)  
             return res;  
         recursion(res,root);  
         return res;  
    }  
     public void recursion(ArrayList<Integer> res, TreeNode root)  
     {  
         if(root.left != null)  
         {  
             recursion(res,root.left);  
         }  
         if(root.right != null)  
         {  
             recursion(res,root.right);  
         }  
         res.add(root.val);  
     }  
}