Binary Tree Inorder Traversal

Binary Tree Inorder Traversal

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?

先贴递归实现的吧

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     List<Integer> list = new ArrayList<Integer>();
12     public List<Integer> inorderTraversal(TreeNode root) {
13         if(null != root){
14             inorderTraversal(root.left);
15             list.add(root.val);
16             inorderTraversal(root.right);
17
18
19         }
20         return list;
21     }
22 }

这里要求用非递归实现，我用的堆栈

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public List<Integer> inorderTraversal(TreeNode root) {
12         List<Integer> result = new ArrayList<Integer>();
13         Stack<TreeNode> stack = new Stack<TreeNode>();
14         if(null == root)
15             return result;
16
17         do{
18             while(null != root){
19                 stack.push(root);
20                 root = root.left;
21             }//左子树入栈
22             TreeNode temp = stack.pop();//出栈
23             result.add(temp.val);
24             root = temp.right;
25             if(null != root)
26             {
27                 stack.push(root);
28                 root = root.left;
29             }
30         }while(!stack.isEmpty());
31
32         return result;
33     }
34 }