题目描述：
LeetCode 545. Boundary of Binary Tree
Given a binary tree, return the values of its boundary in anti-clockwise direction starting from root. Boundary includes left boundary, leaves, and right boundary in order without duplicate nodes.
Left boundary is defined as the path from root to the left-most node. Right boundary is defined as the path from root to the right-most node. If the root doesn't have left subtree or right subtree, then the root itself is left boundary or right boundary. Note this definition only applies to the input binary tree, and not applies to any subtrees.
The left-most node is defined as a leaf node you could reach when you always firstly travel to the left subtree if exists. If not, travel to the right subtree. Repeat until you reach a leaf node.
The right-most node is also defined by the same way with left and right exchanged.
Example 1
Input:
1
\
2
/ \
3 4
Ouput:
[1, 3, 4, 2]
Explanation:
The root doesn't have left subtree, so the root itself is left boundary.
The leaves are node 3 and 4.
The right boundary are node 1,2,4. Note the anti-clockwise direction means you should output reversed right boundary.
So order them in anti-clockwise without duplicates and we have [1,3,4,2].
Example 2
Input:
____1_____
/ \
2 3
/ \ /
4 5 6
/ \ / \
7 8 9 10
Ouput:
[1,2,4,7,8,9,10,6,3]
Explanation:
The left boundary are node 1,2,4. (4 is the left-most node according to definition)
The leaves are node 4,7,8,9,10.
The right boundary are node 1,3,6,10. (10 is the right-most node).
So order them in anti-clockwise without duplicate nodes we have [1,2,4,7,8,9,10,6,3].
题目大意：
给定二叉树，逆时针输出二叉树的边界。边界包括左边界、叶子节点和右边界。
左边界是指从根出发到最左侧节点经过的路径。右边界是指从根出发到最右侧节点经过的路径。
如果根节点不包含左子树或者右子树，则对应的边界不存在。注意此定义是指整棵二叉树，不包含子树。
最左侧节点是指从根节点出发尽量向左走，如果不能则向右走，到达的叶子结点。
最右侧节点定义参考最左侧节点，左右互换即可。
解题思路：
左边界、右边界根据题意求解。叶子节点通过先序遍历得到。
Python代码：
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def boundaryOfBinaryTree(self, root):
"""
:type root: TreeNode
:rtype: List[int]
"""
if not root: return []
if not root.left and not root.right: return [root.val]
leaves = []
def traverse(root):
if not root.left and not root.right:
leaves.append(root)
if root.left:
traverse(root.left)
if root.right:
traverse(root.right)
traverse(root)
left = []
node = root
while node and node != leaves[0]:
left.append(node)
if node.left: node = node.left
else: node = node.right
right = []
node = root
while node and node != leaves[-1]:
right.append(node)
if node.right: node = node.right
else: node = node.left
left = left[1:] if root.left else []
right = right[1:] if root.right else []
return [node.val for node in [root] + left + leaves + right[::-1]]
本文链接：http://bookshadow.com/weblog/2017/03/26/leetcode-boundary-of-binary-tree/
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