Given the root of a binary tree, return the inorder traversal of its nodes' values.
Example 1:
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Input: root = [1,null,2,3]
Output: [1,3,2]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Example 4:
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Input: root = [1,2]
Output: [2,1]
Example 5:
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Input: root = [1,null,2]
Output: [1,2]
Constraints:
- The number of nodes in the tree is in the range
[0, 100].
- -100 <=
Node.val <= 100
Follow up: Recursive solution is trivial, could you do it iteratively?
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# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
''' Recursive '''
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
def inorder(node: Optional[TreeNode], lst: List[int]):
if not node:
return
inorder(node.left, lst)
lst.append(node.val)
inorder(node.right, lst)
result = []
inorder(root, result)
return result
''' Iterative '''
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
if root is None:
return []
stack, res = [], []
curr = root
while curr or stack:
while curr:
stack.append(curr)
curr = curr.left
curr = stack.pop()
res.append(curr.val) # Add after all left children
curr = curr.right
return res
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