LeetCode 94 Binary Tree Inorder Traversal

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?

通常，实现二叉树的前序（preorder）、中序（inorder）、后序（postorder）遍历有两个常用的方法：一是递归(recursive)，二是使用栈实现的迭代版本(stack+iterative)。这两种方法都是O(n)的空间复杂度（递归本身占用stack空间或者用户自定义的stack）

Recursive: Runtime: 1
ms  beats 60.97% of javasubmissions.

public List<Integer> inorderTraversal(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if (root == null) return res;
res.addAll(inorderTraversal(root.left));
res.add(root.val);
res.addAll(inorderTraversal(root.right));
return res;
}

Iterative with Stack. Runtime:
2 ms  beats 4.67% of javasubmissions.

public List<Integer> inorderTraversal2(TreeNode root) {
List<Integer> results = new ArrayList<>();
ArrayDeque<TreeNode> stack = new ArrayDeque<>();
TreeNode crt = root;
while (!stack.isEmpty() || crt != null) {
if (crt != null) {
stack.push(crt);
crt = crt.left;
} else {
crt = stack.pop();
results.add(crt.val);
crt = crt.right;
}
}
return results;
}

Morris Traversal（threaded binary tree）方法遍历二叉树（非递归，不用栈，O(1)空间）中序遍历，时间复杂度 O(n).

Recursive: Runtime: 1
ms  beats 60.97% of javasubmissions.

public List<Integer> inorderTraversal3(TreeNode root) {
List<Integer> results = new ArrayList<>();
TreeNode crt = root;
while (crt != null) {
if (crt.left == null) {
results.add(crt.val);
crt = crt.right;
} else {
TreeNode temp = crt.left;/* 查找前驱 */
while (temp.right != null && temp.right != crt)
temp = temp.right;

if (temp.right == null) {/* 还没线索化，则建立线索 */
temp.right = crt;
crt = crt.left;
} else {/* 已经线索化，则访问节点，并删除线索 */
temp.right = null;
results.add(crt.val);
crt = crt.right;
}
}
}
return results;
}

参考http://www.cnblogs.com/AnnieKim/archive/2013/06/15/MorrisTraversal.html