144. Binary Tree Preorder Traver

2019-06-08  本文已影响0人  云外雁行斜丶

144. Binary Tree Preorder Traversal

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

Example:

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

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

First

1 Stack

根据栈后进先出的原则,先将右节点放入栈中,然后将左节点放入栈中,这样永远先出来
左节点,然后出来右节点。

class Solution(object):
    def preorderTraversal(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.right:
                stack.append(node.right)
            if node.left:
                stack.append(node.left)
        return ans

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.ans.append(root.val)
        self.dfs(root.left)
        self.dfs(root.right)
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