LeetCode Binary Tree Inorder Traversal

原题链接在这里：https://leetcode.com/problems/binary-tree-inorder-traversal/#

本题与Binary Tree Preorder Traversal相呼应。可以分别采用Recursion, Iteration 和 Morris Traversal 三种方法。

Method 1: Recursion

Recursion是traversal时最容易想到，并且好写的方法。

Time O(n), Space O(logn).

AC Java:

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

Method 2: Iteration + Stack

Iteration 时基本就是利用stack的特性体现出recursion的方法。压栈的顺序要注意，一直压left，知道null，再pop，移动root到pop出来的right。循环直到点为空并且stack也为空。

Time O(n), Space O(logn).

AC Java:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {   //Method 2: Iteration   public List<Integer> inorderTraversal(TreeNode root) {       List<Integer> ls = new ArrayList<Integer>();       Stack<TreeNode> stk = new Stack<TreeNode>();       while(root!=null || !stk.empty()){           if(root!=null){               stk.push(root);               root = root.left;           }else{               root = stk.pop();               ls.add(root.val);               root = root.right;           }       }       return ls;   } }

Method 3:

Morris Traversal 算法，参见了Morris Traversal这篇文章。

Time O(n), Space O(1).利用 constant space traverse。

基本思路就是：

1. 看当前点cur的left是否为null，若是null，add当前节点，cur = cur.right.

2. cur.left 不为 null，先在左子树中找到cur 的predecessor，然后分两种情况考虑：

a. predecessor.right = null，那么predecessor.right 指向 cur; cur = cur.left.

b. predecessor.right = cur, 那么先add cur，然后把predecessor.right 改回null，恢复树的结构，然后cur = cur.right.

Note: 1. while()里的条件和下面的if()条件是相呼应的. e.g while()里若写a.next != null, 下面的if()里就应该写a.next 的先关特性。

2. 代码看起来是两个while循环嵌套，感觉是Time 是O(n*logn)，其实每个边最多被访问2次，edge的上限是n-1，所以Time is O(n).

AC Java:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {   //Method 3: Morris Traversal   public List<Integer> inorderTraversal(TreeNode root) {       List<Integer> ls = new ArrayList<Integer>();       TreeNode cur = root;       TreeNode pre = null;       while(cur != null){           if(cur.left == null){               ls.add(cur.val);               cur = cur.right;           }           else{               //Left node is not null, find predcessor                pre = cur.left;               while(pre.right!=null && pre.right!=cur){ //error                   pre = pre.right;               }               if(pre.right == null){                   pre.right = cur;                   cur = cur.left;               }else{                   ls.add(cur.val);                   pre.right = null;                   cur = cur.right;               }           }       }       return ls;   }}