Leetcode_94 Binary Tree Inorder Traversal

祭夏--

于 2015-07-14 14:28:42 发布

阅读量353

点赞数

分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/u014773478/article/details/46877591

版权

LeetCode 专栏收录该内容

33 篇文章

订阅专栏

本文介绍了一种实现二叉树中序遍历的方法，包括递归与非递归两种方式。递归方法直接利用函数调用实现左子树-根节点-右子树的遍历顺序；非递归方法则借助栈来模拟这一过程。

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},
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?

inorder traversal: left child -> parent -> right child

List<Integer> list = new ArrayList<Integer>();
    //recursive 
     public List<Integer> inorderTraversal(TreeNode root) {
            if (root != null) {
                inorder(root);
            }
            return list;
        }
     public void inorder(TreeNode treeNode){
         if (treeNode.left !=null) {
            inorder(treeNode.left);
        }
         list.add(treeNode.val);
         if (treeNode.right != null) {
            inorder(treeNode.right);
        }
     }
     //non-recursive 
     public List<Integer> inorderTraversal1(TreeNode root) {
         Stack<TreeNode> stack = new Stack<TreeNode>();
         TreeNode pNode = root;
         if (root ==null) {
            return list;
        }
         while (pNode !=null || !stack.isEmpty()) {
            if (pNode!=null) {
                stack.push(pNode);
                pNode = pNode.left;

            }else{
                TreeNode temp = stack.pop();
                list.add(temp.val);
                pNode = temp.right;
            }
        }
         return list;
     }