【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?

 

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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