Leetcode 98、102与107
二叉树遍历
二叉树遍历共有前序、中序、后序与层次四种遍历，其中前三种以根出现的位置区分：
前序遍历:根->左->右
中序遍历:左->根->右
后序遍历:左->右->根
而层次是从根到子节点最后到叶子结点的层序遍历方法，具体以leetcode 102和107为例。
Leetcode 102
- 问题描述如下：
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
-
思路：我们用一个队列保存头层结点，遍历一层，一边取值一边添加子结点，取完即可。
-
代码如下：
class Solution:
def levelOrder(self, root):
if not root: return []
res = []
queue = [root]
while queue:
tmp = []
size = len(queue)
while size > 0:
node = queue.pop(0)
tmp.append(node.val)
if node.left: queue.append(node.left)
if node.right: queue.append(node.right)
size -= 1
res.append(tmp)
return res
Leetcode 107
- 问题描述如下：
Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
- 思路：与102一致，最后逆向输出。
- 代码如下：
class Solution:
def levelOrderBottom(self, root):
if not root: return []
ans = []
queue = [root]
while queue:
size = len(queue)
tmp = []
while size > 0:
node = queue.pop(0)
tmp.append(node.val)
if node.left: queue.append(node.left)
if node.right: queue.append(node.right)
size -= 1
ans.append(tmp)
return ans[::-1]
Leetcode 98
Given a binary tree, determine if it is a valid >binary search tree (BST).
Assume a BST is defined as follows:
The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than the node’s key.
Both the left and right subtrees must also be binary >search trees.
Example 1:
Input:
2
/ \
1 3
Output: true
Example 2:
5
/ \
1 4
/ \
3 6
Output: false
- 思路如下：用中序遍历，用一个标签记录上一个结点，如果上一个结点的值大于等于正在遍历的结点，则返回False。
- 代码如下:
class Solution:
def isValidBST(self, root):
if root is None:
return True
stack = []
pre = None
while stack or root:
while root:
stack.append(root)
root = root.left
root = stack.pop()
if pre and pre.val >= root.val: return False
pre = root
root = root.right
return True
