def buildTreeFromPreIn(self,preorder,pB,pE,inorder,iB,iE):
if pB > pE or iB > iE:
return None
if pB == pE:
return TreeNode(preorder[pE])
root = TreeNode(preorder[pB])
mid = 0
for i in range(iB,iE+1):
if inorder[i] == preorder[pB]:
mid = i
break
leftSize = mid-iB
root.left = self.buildTreeFromPreIn(preorder,pB+1,pB+leftSize,inorder,iB,mid-1)
root.right = self.buildTreeFromPreIn(preorder,pB+leftSize+1,pE,inorder,mid+1,iE)
return root
def connect(self, root):
"""
这里给的图确定是一个完全二叉树，但是不是满二叉树。
:type root: TreeLinkNode
:rtype: nothing
"""
if not root:
return
tempRoot = root
nextLineRoot = tempRoot
while tempRoot:
nextLineRoot = None
nextLineRootFound = False
guard = TreeNode(-1)
# 找到当前行的第一个儿子节点 nextLineRoot
while tempRoot:
if tempRoot.left:
if not nextLineRootFound:
nextLineRoot = tempRoot.left
nextLineRootFound = True
guard.next = tempRoot.left
guard = guard.next
if tempRoot.right:
if not nextLineRootFound:
nextLineRoot = tempRoot.right
nextLineRootFound = True
guard.next = tempRoot.right
guard = guard.next
tempRoot = tempRoot.next
if not nextLineRoot: return
tempRoot = nextLineRoot
def dfs(self, begin, end):
"""
从begin到end，生成BST
:param begin:int
:param end: int
:return: List[TreeNode]
"""
ans = []
if begin > end:
ans.append(None)
return ans
if begin == end:
t = TreeNode(begin)
ans.append(t)
return ans
if begin + 1 == end:
t1 = TreeNode(begin)
t1.right = TreeNode(end)
ans.append(t1)
t2 = TreeNode(end)
t2.left = TreeNode(begin)
ans.append(t2)
return ans
for i in range(begin, end + 1):
leftTreeSet = self.dfs(begin, i - 1)
rightTreeSet = self.dfs(i + 1, end)
for nodeL in leftTreeSet:
for nodeR in rightTreeSet:
ro = TreeNode(i)
ro.left = nodeL
ro.right = nodeR
ans.append(ro)
return ans
def buildTreeFromInPost(self, inorder, iB, iE, postorder, pB, pE):
"""
根据中根和后根遍历构建二叉树。后跟遍历的最后一个是根。
:param inorder: List[int]
:param iB:int
:param iE:int
:param postorder: List[int]
:param pB: int
:param pE:int
:return:TreeNode
"""
if iB > iE or pB > pE: return None
if iB == iE:
return TreeNode(inorder[iE])
root = TreeNode(postorder[pE])
mid = inorder.index(postorder[pE])
leftLen = mid - iB
root.left = self.buildTreeFromInPost(inorder, iB, mid - 1, postorder,
pB, pB + leftLen - 1)
root.right = self.buildTreeFromInPost(inorder, mid + 1, iE, postorder,
pB + leftLen, pE - 1)
return root
def buildBST(self,begin,end):
"""
按照中根遍历的顺序构建BST
:param head: ListNode
:param begin: int
:param end: int
:return: TreeNode
"""
if begin > end:
return None
mid = (begin+end)//2
leftTree = self.buildBST(begin,mid-1)
root = TreeNode(self.node.val)
root.left = leftTree
self.node = self.node.next
root.right = self.buildBST(mid+1,end)
return root
