94. Binary Tree Inorder Traversal java solutions

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

For example:
Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

 

return [1,3,2].

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

 

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 1 /**
 2  * Definition for a binary tree node.
 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         ArrayList<Integer> ans = new ArrayList<Integer>();
13         Stack<TreeNode> s = new Stack<TreeNode>();
14 
15         if(root == null) return ans;
16         s.push(root);
17         while(!s.isEmpty()){
18             TreeNode node = s.peek();
19             if(node.left != null){
20                 s.push(node.left);
21                 node.left = null;//原先这里没有设置为null一直内存超出限制，过不去。不知道为什么，
22             }else{
23                 ans.add(node.val);
24                 s.pop();
25                 if(node.right != null){
26                     s.push(node.right);
27                 }
28             }
29         }
30         
31         return ans;
32     }
33 }

 

posted @ 2016-06-23 10:48  Miller1991  阅读(125)  评论(0)  编辑  收藏  举报