Leetcode 94. Binary Tree Inorder Traversal

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].
题意：中序遍历二叉树
题解：用递归的方法很容易求解，但是对于大部分计算机而言，递归比循环慢得多，所以这里给出一种运用栈的方法。
Solution1： recrusion

public class Solution {
 private List<Integer> list=new ArrayList<Integer>();
 public List<Integer> inorderTraversal(TreeNode root) {
           if(root!=null)
           {
            inorderTraversal(root.left);
            list.add(root.val);
            inorderTraversal(root.right);
           }
           return list;
          }
        }

Solution2：Stack&&Iteration

public class Solution {
 private List<Integer> list=new ArrayList<Integer>();
 public List<Integer> inorderTraversal(TreeNode root) {
       TreeNode x=root;
       Stack<TreeNode>s=new Stack<TreeNode>();
       while(x!=null||!s.isEmpty())
       {
           while(x!=null)
           {
               s.push(x);
               x=x.left;
           }
           x=s.peek();
           s.pop();
           list.add(x.val);
           x=x.right;
       }
       return list;
    }
}