Binary tree inorder traversal

Time: O(N); Space: O(1); medium

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

Example 1:

  1
 / \
2   3

Input：root = {TreeNode} [1,2,3]

Output：{TreeNode} [2,1,3]

Example 2:

1
 \
  2
 /
3

Input: root = {TreeNode} [1,#,2,3]

Output: {TreeNode} [1,3,2]

Note:

  • Recursive solution is trivial, could you do it iteratively?

[1]:

class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

1. Morris Traversal Solution

[2]:

class Solution1(object):
    '''
    Time: O(N)
    Space: O(1)
    '''
    def inorderTraversal(self, root):
        '''
        :type root: TreeNode
        :rtype: List[int]
        '''
        result, curr = [], root
        while curr:
            if curr.left is None:
                result.append(curr.val)
                curr = curr.right
            else:
                node = curr.left
                while node.right and node.right != curr:
                    node = node.right

                if node.right is None:
                    node.right = curr
                    curr = curr.left
                else:
                    result.append(curr.val)
                    node.right = None
                    curr = curr.right

        return result

[3]:

s = Solution1()

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
assert s.inorderTraversal(root) == [2, 1, 3]

root = TreeNode(1)
root.right = TreeNode(2)
root.right.left = TreeNode(3)
assert s.inorderTraversal(root) == [1, 3, 2]

2. Stack Solution

[4]:

class Solution2(object):
    '''
    Time: O(N)
    Space: O(H)
    '''
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        result, stack = [], [(root, False)]
        while stack:
            root, is_visited = stack.pop()
            if root is None:
                continue
            if is_visited:
                result.append(root.val)
            else:
                stack.append((root.right, False))
                stack.append((root, True))
                stack.append((root.left, False))
        return result

[5]:

s = Solution2()

root = TreeNode(1)
root.left = TreeNode(2)
root.right = TreeNode(3)
assert s.inorderTraversal(root) == [2, 1, 3]

root = TreeNode(1)
root.right = TreeNode(2)
root.right.left = TreeNode(3)
assert s.inorderTraversal(root) == [1, 3, 2]