给定一个二叉树的根节点 root ，返回它的 中序 遍历。
image.png

示例 1：
输入：root = [1,null,2,3]
输出：[1,3,2]

递归：

  1. /**
  2.  * Definition for a binary tree node.
  3.  * public class TreeNode {
  4.  *     int val;
  5.  *     TreeNode left;
  6.  *     TreeNode right;
  7.  *     TreeNode() {}
  8.  *     TreeNode(int val) { this.val = val; }
  9.  *     TreeNode(int val, TreeNode left, TreeNode right) {
  10.  *         this.val = val;
  11.  *         this.left = left;
  12.  *         this.right = right;
  13.  *     }
  14.  * }
  15.  */
  16. class Solution {
  17.     public List<Integer> inorderTraversal(TreeNode root) {
  18.         List<Integer> res = new ArrayList<>();
  19.         inorderTraversal(root, res);
  20.         return res;
  22.     }
  23.     public void inorderTraversal(TreeNode node, List<Integer> res){
  24.         if(node == null){
  25.             return;
  26.         }
  27.         //中序遍历 left->root->right
  28.         inorderTraversal(node.left, res);
  29.         keys.add(node.val);
  30.         inorderTraversal(node.right, res);
  31.     }
  32. }

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<>();
        Stack<TreeNode> stack = new Stack<>();
        if(root == null){
            return res;
        }
        TreeNode cur = root;
        while(cur!=null || !stack.isEmpty()){
           if(cur!=null){
               stack.push(cur);
             cur = cur.left;
           }else{
                   cur = stack.pop();
               res.add(cur.val);
               cur = cur.left;
           }

        }
  return res;
    }
}