145. Binary Tree Postorder Trave

145. Binary Tree Postorder Traversal

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

Example:

Input: [1,null,2,3]
   1
    \
     2
    /
   3

Output: [3,2,1]

Follow up: Recursive solution is trivial, could you do it iteratively?

First

Stack

同样，根据树的后续遍历的规律，然后逐个将子节点加入栈

class Solution(object):
    def postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        stack = [root]
        ans = []
        while stack:
            node = stack.pop()
            ans.append(node.val)
            if node.left:
                stack.append(node.left)
            if node.right:
                stack.append(node.right)
        return ans[::-1]

Recursion

class Solution(object):
    def postorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        self.ans = []
        self.dfs(root)
        return self.ans

    def dfs(self, root):
        if root is None:
            return
        self.dfs(root.left)
        self.dfs(root.right)
        self.ans.append(root.val)