654. 最大二叉树
题目
思路与解法
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def constructMaximumBinaryTree(self, nums: List[int]) -> Optional[TreeNode]:
if not nums:
return None
max_val = max(nums)
max_idx = nums.index(max_val)
root = TreeNode(max_val)
root.left = self.constructMaximumBinaryTree(nums[:max_idx])
root.right = self.constructMaximumBinaryTree(nums[max_idx+1:])
return root617.合并二叉树
题目
思路与解法
第一想法： 通过层序遍历来合并
一遍过，哈哈，虽然肯定有可以优化的地方
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def mergeTrees(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> Optional[TreeNode]:
if not root1:
return root2
if not root2:
return root1
from collections import deque
queue1 = deque()
queue2 = deque()
queue1.append(root1)
queue2.append(root2)
while queue1 or queue2:
cur1 = queue1.popleft()
cur2 = queue2.popleft()
cur_new_val = cur1.val + cur2.val
cur1.val = cur_new_val
if not cur1.left and cur2.left:
cur1.left = TreeNode(0)
elif cur1.left and not cur2.left:
cur2.left = TreeNode(0)
if not cur1.right and cur2.right:
cur1.right = TreeNode(0)
elif cur1.right and not cur2.right:
cur2.right = TreeNode(0)
if cur1.left:
queue1.append(cur1.left)
queue2.append(cur2.left)
if cur1.right:
queue1.append(cur1.right)
queue2.append(cur2.right)
return root1700.二叉搜索树中的搜索
题目
思路与解法
第一想法：
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def searchBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]:
val_node = self.find_root(root, val)
return val_node
def find_root(self, node, val):
if not node:
return None
if node.val == val:
return node
elif node.val > val:
return self.find_root(node.left, val)
elif node.val < val:
return self.find_root(node.right, val)98.验证二叉搜索树
题目
思路与解法
第一想法： 判断每个结点作为root结点构成的子树是不是二叉搜索树，如下。但是有坑，光判断每个子数是不够的，因为子树中结点可能不满足子树外的条件
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isValidBST(self, root: Optional[TreeNode]) -> bool:
if not root:
return True
if root.left and root.left.val >= root.val:
return False
if root.right and root.right.val <= root.val:
return False
return self.isValidBST(root.left) and self.isValidBST(root.right)如下就不满足条件，
思考后的想法： 中序遍历后，挨个判断是不是递增的。发现carl也是这么做的，只是他有几种方法。
python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def isValidBST(self, root: Optional[TreeNode]) -> bool:
self.nodes = []
self.traversal(root)
for i in range(len(self.nodes)-1):
if self.nodes[i] >= self.nodes[i+1]:
return False
return True
def traversal(self, root):
if not root:
return None
self.traversal(root.left)
self.nodes.append(root.val)
self.traversal(root.right)