Merge pull request #329 from z80160280/master

Update 0222.完全二叉树的节点个数.md
This commit is contained in:
Carl Sun
2021-06-06 19:57:25 +08:00
committed by GitHub
2 changed files with 121 additions and 0 deletions

View File

@ -498,6 +498,62 @@ class Solution {
Python
> 递归法:
```python
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
return True if self.getDepth(root) != -1 else False
#返回以该节点为根节点的二叉树的高度,如果不是二叉搜索树了则返回-1
def getDepth(self, node):
if not node:
return 0
leftDepth = self.getDepth(node.left)
if leftDepth == -1: return -1 #说明左子树已经不是二叉平衡树
rightDepth = self.getDepth(node.right)
if rightDepth == -1: return -1 #说明右子树已经不是二叉平衡树
return -1 if abs(leftDepth - rightDepth)>1 else 1 + max(leftDepth, rightDepth)
```
> 迭代法:
```python
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
st = []
if not root:
return True
st.append(root)
while st:
node = st.pop() #中
if abs(self.getDepth(node.left) - self.getDepth(node.right)) > 1:
return False
if node.right:
st.append(node.right) #右(空节点不入栈)
if node.left:
st.append(node.left) #左(空节点不入栈)
return True
def getDepth(self, cur):
st = []
if cur:
st.append(cur)
depth = 0
result = 0
while st:
node = st.pop()
if node:
st.append(node) #中
st.append(None)
depth += 1
if node.right: st.append(node.right) #右
if node.left: st.append(node.left) #左
else:
node = st.pop()
depth -= 1
result = max(result, depth)
return result
```
Go
```Go

View File

@ -240,6 +240,71 @@ class Solution {
Python
> 递归法:
```python
class Solution:
def countNodes(self, root: TreeNode) -> int:
return self.getNodesNum(root)
def getNodesNum(self, cur):
if not cur:
return 0
leftNum = self.getNodesNum(cur.left) #左
rightNum = self.getNodesNum(cur.right) #右
treeNum = leftNum + rightNum + 1 #中
return treeNum
```
> 递归法:精简版
```python
class Solution:
def countNodes(self, root: TreeNode) -> int:
if not root:
return 0
return 1 + self.countNodes(root.left) + self.countNodes(root.right)
```
> 迭代法:
```python
import collections
class Solution:
def countNodes(self, root: TreeNode) -> int:
queue = collections.deque()
if root:
queue.append(root)
result = 0
while queue:
size = len(queue)
for i in range(size):
node = queue.popleft()
result += 1 #记录节点数量
if node.left:
queue.append(node.left)
if node.right:
queue.append(node.right)
return result
```
> 完全二叉树
```python
class Solution:
def countNodes(self, root: TreeNode) -> int:
if not root:
return 0
left = root.left
right = root.right
leftHeight = 0 #这里初始为0是有目的的为了下面求指数方便
rightHeight = 0
while left: #求左子树深度
left = left.left
leftHeight += 1
while right: #求右子树深度
right = right.right
rightHeight += 1
if leftHeight == rightHeight:
return (2 << leftHeight) - 1 #注意(2<<1) 相当于2^2所以leftHeight初始为0
return self.countNodes(root.left) + self.countNodes(root.right) + 1
```
Go