华子目录
平衡二叉树
思路
- 什么是
平衡二叉树：左子树的高度与右子树的高度相差不超过1 isBalanced用来判断是否为平衡二叉树，如果当前树不是平衡二叉树，直接返回最终结果False。如果当前树为平衡二叉树，就再判断其左右子树，左右子树也必须为平衡二叉树cal_depth用来求一个子树的最大深度，这是之前讲过的内容（求最大深度、最小深度、节点个数，都是基础内容）
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 cal_depth(self,root):
cur = root
if not cur:
return 0
leftHeight = self.cal_depth(cur.left)
rightHeight = self.cal_depth(cur.right)
height = 1 + max(leftHeight, rightHeight)
return height
def isBalanced(self, root: Optional[TreeNode]) -> bool:
if not root:
return True
left_depth = self.cal_depth(root.left) # 求出左子树的深度
right_depth = self.cal_depth(root.right) # 求出右子树的深度
if abs(left_depth - right_depth) > 1: # 相差
return False
return self.isBalanced(root.left) and self.isBalanced(root.right)二叉树的所有路径
思路
- 这道题是使用
回溯来求解，第一次接触回溯法，多看看视频来理解。
具体写法
递归的写法里传入三个参数：（1）当前处理的节点；（2）当前的路径；（3）总的结果集递归的终止条件：当节点为叶子节点时，把当前路径加入结果集递归的单层处理：当node有左节点时，递归处理左节点，处理结束后回溯。当node有右节点时，递归处理右节点，处理结束后回溯
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 travel(self, node, path, res):
if not node.left and not node.right: # node为叶子节点
path = [str(x) for x in path]
res.append('->'.join(path)) # 将某一条路径加入到总的结果集当中
if node.left: # 左有值
path.append(node.left.val)
self.travel(node.left, path, res)
path.pop() # 某一条路线递归结束后，弹出
if node.right: # 右有值
path.append(node.right.val)
self.travel(node.right, path, res)
path.pop() # 某一条路径递归结束后，弹出
def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
cur = root
if not cur:
return []
res = [] # 记录所有路径
path = [cur.val] # 记录当前路径
self.travel(cur,path, res)
return res左叶子之和
思路
- 使用
后序遍历
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 sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
if not root: # 终止条件，当前节点为空
return 0
if not root.left and not root.right: # 终止条件，当前节点是叶子节点
return 0
left_num = self.sumOfLeftLeaves(root.left)
right_num = self.sumOfLeftLeaves(root.right)
cur_num = 0
if root.left and not root.left.left and not root.left.right: # # root.left 是左叶子
cur_num = root.left.val # 记录左叶子的值
return cur_num + left_num + right_num # 三部分的汇总完全二叉树的节点个数
思路
- 把
完全二叉树当成二叉树去计算，使用后序遍历去计算
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 countNodes(self, root: Optional[TreeNode]) -> int:
cur = root
if not cur:
return 0
left_num = self.countNodes(cur.left)
right_num = self.countNodes(cur.right)
result = left_num + right_num + 1 # 核心
return result