94. Binary Tree Inorder Traversal
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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]
Example 2:
Input: root = []
Output: []
Example 3:
Input: root = [1]
Output: [1]
Thoughts
Don’t understand what needed. Why:
1-null-2-3becomes1-3-2[1,2,5,7,8,9,10]becomes[7,2,8,1,9,5,10]
In 1-null-2-3 1 becomes the first because we loop to its left node which is null, then come back and first value here is 1.
First accepted
# 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
class Solution:
def inorderTraversal(self, root: Optional[TreeNode]) -> List[int]:
# add all left, then add right
def get_child(head):
if head:
get_child(head.left)
result.append(head.val)
get_child(head.right)
result = []
get_child(root)
return result
Better solution
Morris Traversal
