94-binary-tree-Inorder-traversal

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

Example 1:

Input: root = [1,null,2,3]

Output: [1,3,2]

Explanation:

Example 2:

Input: root = [1,2,3,4,5,null,8,null,null,6,7,9]

Output: [4,2,6,5,7,1,3,9,8]

Explanation:

Example 3:

Input: root = []

Output: []

Example 4:

Input: root = [1]

Output: [1]

Recursion solution：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: Optional[TreeNode]
        :rtype: List[int]
        """
        res = []
        def dfs(parent):
            if not parent:
                return
            if parent.left:
                dfs(parent.left)
            res.append(parent.val)
            if parent.right:
                dfs(parent.right)
        dfs(root)
        return res

        

Iteration solution：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: Optional[TreeNode]
        :rtype: List[int]
        """
        res = []
        stack = []
        curr = root
        while curr or stack:
            if curr:
                stack.append(curr)
                curr = curr.left
            else:
                curr = stack.pop()
                res.append(curr.val)
                curr = curr.right
        return res