PU Binary Tree Inorder Traversal

Jan 01, 1970

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?

Python Solution 1:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        res = []
        def inorder(root):
            if not root: return
            inorder(root.left)
            res.append(root.val)
            inorder(root.right)
        inorder(root)
        return res

Python Solution 2:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def inorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if not root: return []
        stack = [root]
        res = []
        while (stack): 
            node = stack[-1]
            if node.left: 
                stack.append(node.left)
                node.left = None
                continue
            stack.pop()
            res.append(node.val)
            if node.right:
                stack.append(node.right)
        return res

Summary:

  • nothing to say

LeetCode: 94. Binary Tree Inorder Traversal