1、求树的所有路径和:
class Solution_iterative:
def hasPathSum(self, root, target):
stack = [(root, 0)]
while stack:
node, path = stack.pop()
if node:
if node.left is None and node.right is None and path + node.val == target:
return True
stack.extend([(node.left, path + node.val), (node.right, path + node.val)])
return False
2、交换左右子树
def invertTree(self, root):
stack = [root]
while stack:
node = stack.pop()
if node:
node.left, node.right = node.right, node.left
stack += node.left, node.right
return root
3、求树的每层平均值
def BFS（root):
if not root:
return 0
prev,res=[root],[]
while prev:
cur=[]
#给结果添加当前层数prev的平均值，prev为当前所有的值
val=sum([node.val for node in prev])/float(len(prev))
res.append(val)
#该循环主要是cur添加prev所有的左右子树，即cur为下一层所有的值
while prev:
node=prev.pop()
if node.left:
cur.append(node.left)
if node.right:
cur.append(node.right)
#将当前层数变为下一层
prev=cur
return res
4、二叉树的层次遍历
def levelOrder(self, root):
"""
:type root: TreeNode
:rtype: List[List[int]]
"""
res=[]
if not root:
return []
prev=[root]
while prev:
temp=[node.val for node in prev]
res.append(temp)
cur=[]
while prev:
node=prev.pop(False)
if node.left:
cur.append(node.left)
if node.right:
cur.append(node.right)
prev=cur
return res
5、二叉树的第二小节点
def findSecondMinimumValue(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if not root:
return -1
prev=[root]
minVal=root.val
secondVal=float('inf')
while prev:
cur=[]
while prev:
node=prev.pop()
if minVal<node.val and node.val<secondVal:
secondVal=node.val
if node.left:
cur.append(node.left)
if node.right:
cur.append(node.right)
prev=cur
return -1 if secondVal==float('inf') else secondVal
6、最长同值路径：
class Solution_iterative:
def longestUnivaluePath(self, root):
postorder = [(0, root, None)]
count = 0
d = {None: 0}
while postorder:
seen, node, parent = postorder.pop()
if node is None:
continue
if not seen:
postorder.append((1, node, parent))
postorder.append((0, node.right, node.val))
postorder.append((0, node.left, node.val))
else:
if node.val == parent:
d[node] = max(d[node.left], d[node.right]) + 1
else:
d[node] = 0
count = max(count, d[node.left] + d[node.right])
return count
7、判断一棵树是否为高度平衡二叉树
def isBalanced(self, root):
stack = [(0, root)]
depth = {None: 0}
while stack:
seen, node = stack.pop()
if node is None:
continue
if not seen:
stack.extend([(1, node), (0, node.right), (0, node.left)])
else:
if abs(depth[node.left] - depth[node.right]) > 1:
return False
depth[node] = max(depth[node.left], depth[node.right]) + 1
return True
