原题地址:http://oj.leetcode.com/problems/binary-tree-inorder-traversal/
题意:二叉树的中序遍历。
解题思路:这道题用递归解不难,所以应该考察的是非递归求解二叉树的中序遍历。我们使用一个栈来解决问题。比如一颗二叉树为{1,2,3,4,5,6,7},第一层为{1},第二层为{2,3},第三层为{4,5,6,7}。那么我们依次存储左子树的根节点,那么入栈顺序为:1,2,4。由于4的左子树为空,所以开始出栈。4出栈,检查4的右子树为空,继续出栈,2出栈后,2的右子树非空,5入栈,5的左子树为空,5出栈,由于5的右子树为空,继续出栈,1出栈后,发现1的右子树非空,3入栈。。。依次类推,就完成了非递归(迭代)解法。
代码:
# Definition for a binary tree node
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = Noneclass Solution:
# @param root, a tree node
# @return a list of integers
def iterative_inorder(self, root, list):
stack = []
while root or stack:
if root:
stack.append(root)
root = root.left
else:
root = stack.pop()
list.append(root.val)
root = root.right
return listdef recursive_inorder(self, root, list):
if root:
self.inorder(root.left, list)
list.append(root.val)
self.inorder(root.right, list)def inorderTraversal(self, root):
list = []
self.iterative_inorder(root, list)
return list
[leetcode]Binary Tree Inorder Traversal @ Python