给定一个二叉树,返回它的中序 遍历。

示例:

输入: [1,null,2,3] 1 \ 2 / 3 输出: [1,3,2]

进阶: 递归算法很简单,你可以通过迭代算法完成吗?


 

递归:

  public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();

        if (root == null) {
            return result;
        }

        result.addAll(inorderTraversal(root.left));

        result.add(root.val);

        result.addAll(inorderTraversal(root.right));


        return result;

    }

非递归:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
     public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> result = new ArrayList<>();

        if (root == null) {
            return result;
        }

        Stack<TreeNode> stack = new Stack<>();
        TreeNode curNode = root;

        while (!stack.isEmpty() || curNode != null) {

            if (curNode != null) {
                stack.push(curNode);
                curNode = curNode.left;
            } else {
                curNode = stack.pop();
                result.add(curNode.val);
                curNode = curNode.right;
            }
        }

        return result;

    }
}