66. 二叉树的前序遍历

66. 二叉树的前序遍历

中文English

给出一棵二叉树，返回其节点值的前序遍历。

样例

样例 1:

输入：{1,2,3}
输出：[1,2,3]
解释：
   1
  / \
 2   3
它将被序列化为{1,2,3}
前序遍历

样例 2:

输入：{1,#,2,3}
输出：[1,2,3]
解释：
1
 \
  2
 /
3
它将被序列化为{1,#,2,3}
前序遍历

挑战

你能使用非递归实现么？

注意事项

• 首个数据为根节点，后面接着是其左儿子和右儿子节点值，"#"表示不存在该子节点。
• 节点数量不超过20

 

 

输入测试数据 (每行一个参数)如何理解测试数据？

 分治法:

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: A Tree
    @return: Preorder in ArrayList which contains node values.
    """
        
    def preorderTraversal(self, root):
        # write your code here
        if not root: return []
        results = []
        
        left = self.preorderTraversal(root.left)
        right = self.preorderTraversal(root.right)
        
        results.append(root.val)
        results.extend(left)
        results.extend(right)
        
        return results 
        

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: A Tree
    @return: Preorder in ArrayList which contains node values.
    """
        
    def preorderTraversal(self, root):
        # write your code here
        return [root.val] + self.preorderTraversal(root.left) + self.preorderTraversal(root.right) if root else []

 

非递归版本:

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: A Tree
    @return: Preorder in ArrayList which contains node values.
    """
        
    def preorderTraversal(self, root):
        # write your code here
        
        #非递归
        results = []
        stack = [root]
        
        while stack:
            pop_node = stack.pop(0)
            results.append(pop_node.val)
            if pop_node.left:
                stack.append(pop_node.left)
            if pop_node.right:
                stack.append(pop_node.right)
        
        return results