LeetCode 94. Binary Tree Inorder Traversal

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

题目：

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

   1
    
     2
    /
   3

return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

题解：

类似Binary Tree Preorder Traversal, Binary Tree Postorder Traversal.

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 Complexity: 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, 直到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 Complexity: 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> res = new ArrayList<Integer>();
        if(root == null){
            return res;
        }
        Stack<TreeNode> stk = new Stack<TreeNode>();
        while(root != null || !stk.isEmpty()){
            if(root != null){
                stk.push(root);
                root = root.left;
            }else{
                TreeNode tn = stk.pop();
                res.add(tn.val);
                root = tn.right;
            }
        }
        return res;
    }
}

Method 3:

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

Time Complexity: 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, 表明左子树还没遍历过。那么predecessor.right 指向 cur; cur = cur.left.

b. predecessor.right = cur, 表明左子树已经遍历完了。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 {
    public List<Integer> inorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        TreeNode cur = root;
        TreeNode pre = null;
        while(cur != null){
            if(cur.left == null){
                res.add(cur.val);
                cur = cur.right;
            }else{
                //Left node is not null, find predcessor
                pre = cur.left;
                while(pre.right != null && pre.right != cur){
                    pre = pre.right;
                    
                }
                
                if(pre.right == null){
                    pre.right = cur;
                    cur = cur.left;
                }else{
                    res.add(cur.val);
                    pre.right = null;
                    cur = cur.right;
                }
            }
        }
        return res;
    }
}

跟上Convert Binary Search Tree to Sorted Doubly Linked List.