*Binary Tree Inorder Traversal

Posted 2020-08-12 Hygeia

题目：

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?

 

题解：

中序遍历：递归左 处理当前 递归右。

 

解法一：递归代码如下：

 1 public void helper(TreeNode root, ArrayList<Integer> re){
 2         if(root==null)
 3             return;
 4         helper(root.left,re);
 5         re.add(root.val);
 6         helper(root.right,re);
 7     }
 8     public ArrayList<Integer> inorderTraversal(TreeNode root) {
 9         ArrayList<Integer> re = new ArrayList<Integer>();
10         if(root==null)
11             return re;
12         helper(root,re);
13         return re;
14     }

 

解法二：非递归解法

 1 public ArrayList<Integer> inorderTraversal(TreeNode root) {  
 2     ArrayList<Integer> res = new ArrayList<Integer>();  
 3     if(root == null)  
 4         return res;  
 5     LinkedList<TreeNode> stack = new LinkedList<TreeNode>();  
 6     while(root!=null || !stack.isEmpty()){  
 7         if(root!=null){
 8             stack.push(root);
 9             root = root.left; 
10         }else{  
11             root = stack.pop();
12             res.add(root.val); 
13             root = root.right;  
14         }  
15     }  
16     return res;  
17 }

 

reference：http://www.cnblogs.com/springfor/p/3877179.html