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 subtree
     
    
   

    

     

    
    

     

       pre = curr.left;
     
    
   

    

     

    
    

     
 while (pre.right != null) { // find rightmost
     
    
   

    

     

    
    

     

       pre = pre.right;
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

       pre.right = curr; // put cur after the pre node
     
    
   

    

     

    
    

     

       TreeNode temp = curr; // store cur node
     
    
   

    

     

    
    

     

       curr = curr.left; // move cur to the top of the new tree
     
    
   

    

     

    
    

     

       temp.left = null; // original cur left be null, avoid infinite loops
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     
 return res;
     
    
   

    

     

    
    

     

       }
     
    
   

    

     

    
    

     

      }
     
    
  
 

文章来源: fantianzuo.blog.csdn.net，作者：兔老大RabbitMQ，版权归原作者所有，如需转载，请联系作者。

原文链接：fantianzuo.blog.csdn.net/article/details/103299465