方法一：（递归）
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public static void dfs(TreeNode root,List<Integer> res) { if(root==null) { return; } dfs(root.left,res); res.add(root.val); dfs(root.right,res); } public List<Integer> inorderTraversal(TreeNode root) { List<Integer> res=new ArrayList<Integer>(); dfs(root,res); return res; } }方法二：（迭代）
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode() {} * TreeNode(int val) { this.val = val; } * TreeNode(int val, TreeNode left, TreeNode right) { * this.val = val; * this.left = left; * this.right = right; * } * } */ class Solution { public List<Integer> inorderTraversal(TreeNode root) { List<Integer> res=new ArrayList<Integer>(); LinkedList<TreeNode> queue=new LinkedList<TreeNode>(); TreeNode node=root; while(node!=null || !queue.isEmpty()) { while(node!=null) { queue.offer(node); node=node.left; } node=queue.pollLast(); res.add(node.val); node=node.right; } return res; } }方法三：（Morris）
