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

/*** 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 > res = new ArrayList < > ();helper(root, res);return res;}public void helper(TreeNode root, List < Integer > res) {if(root == null)return;helper(root.left, res);res.add(root.val);helper(root.right, res);}
}

压栈

public class Solution {public List < Integer > inorderTraversal(TreeNode root) {List < Integer > res = new ArrayList < > ();Stack < TreeNode > stack = new Stack < > ();TreeNode curr = root;while (curr != null || !stack.isEmpty()) {while (curr != null) {stack.push(curr);curr = curr.left;}curr = stack.pop();res.add(curr.val);curr = curr.right;}return res;}
}

morris

虽然是空间O(1),但是oj并没有测出来效果，依旧空间只超过40%

/*** 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 > res = new ArrayList < > ();TreeNode curr = root;TreeNode pre;while (curr != null) {if (curr.left == null) {res.add(curr.val);curr = curr.right; // move to next right node} else { // has a left subtreepre = curr.left;while (pre.right != null) { // find rightmostpre = pre.right;}pre.right = curr; // put cur after the pre nodeTreeNode temp = curr; // store cur nodecurr = curr.left; // move cur to the top of the new treetemp.left = null; // original cur left be null, avoid infinite loops}}return res;}
}