PU Binary Tree Postorder Traversal

Jan 01, 1970

Given a binary tree, return the postorder traversal of its nodes' values.

For example:

Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [3,2,1].

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 postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        res = []
        def postorder(root):
            if not root: return
            postorder(root.left)
            postorder(root.right)
            res.append(root.val)
        postorder(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 postorderTraversal(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
            if node.right:
                stack.append(node.right)
                node.right = None
                continue
            stack.pop()
            res.append(node.val)
        return res

Summary:

  • nothing to say

LeetCode: 145. Binary Tree Postorder Traversal