144. Binary Tree Preorder Traversal

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

For example:
Given binary tree {1,#,2,3},

   1
    
     2
    /
   3

return [1,2,3].

 该题是对树做前序遍历

下面分别是递归，非递归，分治三种思路的解题结果

#递归写法
class Solution(object):
    
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        result = []
        self.traverse(root,result)
        return result
    def traverse(self,root,result):
        if root is None:
            return
        result.append(root.val)
        self.traverse(root.left,result)
        self.traverse(root.right,result)   

#分治法
class Solution(object):
    
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        result = []
        #分
        left = self.preorderTraversal(root.left)
        right = self.preorderTraversal(root.right)
        #治
        result.append(root.val)
        result.extend(left)
        result.extend(right)
        return result

#非递归写法，用栈模拟
class Solution(object):
    def preorderTraversal(self, root):
        """
        :type root: TreeNode
        :rtype: List[int]
        """
        if root is None:
            return []
        stack = [root]
        preorder = []
        while stack:
            node = stack.pop()
            preorder.append(node.val)
            if(node.right):
                stack.append(node.right)
            if(node.left):
                stack.append(node.left)
        return preorder