【LeetCode】Binary Tree Preorder Traversal

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].

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

---

Solution:

递归解法很简单：

 # Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None

 class Solution:
     # @param {TreeNode} root
     # @return {integer[]}
     def preorderTraversal(self, root):
         if root == None:
             return []
         else:
             return [root.val] + self.preorderTraversal(root.left) + self.preorderTraversal(root.right)

迭代版本也是常规的将递归改成迭代的版本：

用一个栈来模拟递归的过程，注意栈 FILO 的特点。所以，对于当前根，要把右子树先加入栈，然后再把左子树加入栈。

前序遍历的顺序是：根 - 左子树 - 右子树。

代码如下：

 # Definition for a binary tree node.
 # class TreeNode:
 #     def __init__(self, x):
 #         self.val = x
 #         self.left = None
 #         self.right = None

 class Solution:
     # @param {TreeNode} root
     # @return {integer[]}
     def preorderTraversal(self, root):
         stack = []
         path = []
         if root != None:
             stack.append(root)
         while stack != []:
             e = stack.pop()
             path.append(e.val)
             if e.right != None:
                 stack.append(e.right)
             if e.left != None:
                 stack.append(e.left)
         return path