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

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    List<Integer> list = new ArrayList<Integer>();
    public List<Integer> inorderTraversal(TreeNode root) {
        if(null != root){
            inorderTraversal(root.left);
            list.add(root.val);
            inorderTraversal(root.right);
        
            
        }
        return list;
    }
}

这里要求用非递归实现，我用的堆栈

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack<TreeNode>();
        if(null == root)
            return result;

        do{
            while(null != root){
                stack.push(root);
                root = root.left;
            }//左子树入栈
            TreeNode temp = stack.pop();//出栈
            result.add(temp.val);
            root = temp.right;
            if(null != root)
            {
                stack.push(root);
                root = root.left;
            }
        }while(!stack.isEmpty());
        
        return result;
    }
}