【每日一题】【栈】【递归】【遍历】2021年12月1日-94. 二叉树的中序遍历

TreeNode给定一个二叉树的根节点 root ，返回它的 中序 遍历。

 两种方式：递归（先序后序也要掌握）或非递归

/**
 * 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;
 *     }
 * }
 */
 //简单的递归方式
class Solution {
    public List inorderTraversal(TreeNode root) {
        List list = new ArrayList<>();
        inOrder(root, list);
        return list;
    }

    public void inOrder(TreeNode root, List list) {
        if(root == null) {
            return;
        }
        inOrder(root.left, list);
        list.add(root.val);
        inOrder(root.right, list);
    }
}
//非递归方式
class Solution {
    public List inorderTraversal(TreeNode root) {
        List list = new ArrayList<>();
        Stack stack = new Stack<>();
　　　　 //Deque stack = new LinkedList();
        //栈非空是isEmpty()而不是!=null
        while(root != null || !stack.isEmpty()) {
            while(root != null) {
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            list.add(root.val);
            root = root.right;
        }
        return list;
    }
}