【leetcode刷题之路】面试经典150题（5）——二叉树+二叉树层次遍历+二叉搜索树

文章目录

- - 9 二叉树
  - - 9.1 【递归】二叉树的最大深度
    - 9.2 【递归】相同的树
    - 9.3 【递归】翻转二叉树
    - 9.4 【递归】对称二叉树
    - 9.5 【递归】从前序与中序遍历序列构造二叉树
    - 9.6 【递归】从中序与后序遍历序列构造二叉树
    - 9.7 【BFS】填充每个节点的下一个右侧节点指针 II
    - 9.8 【递归】二叉树展开为链表
    - 9.9 【DFS】路径总和
    - 9.10 【DFS】求根节点到叶节点数字之和
    - 9.11 【BFS】【动态规划】二叉树中的最大路径和
    - 9.12 【BFS】二叉搜索树迭代器
    - 9.13 【BFS】完全二叉树的节点个数
    - 9.14 【递归】二叉树的最近公共祖先
  - 10 二叉树层次遍历
  - - 10.1 【DFS】二叉树的右视图
    - 10.2 【BFS】二叉树的层平均值
    - 10.3 【BFS】二叉树的层序遍历
    - 10.4 【BFS】二叉树的锯齿形层序遍历
  - 11 二叉搜索树
  - - 11.1 【BFS】二叉搜索树的最小绝对差
    - 11.2 【BFS】二叉搜索树中第K小的元素
    - 11.3 【BFS】【递归】验证二叉搜索树

9 二叉树

9.1 【递归】二叉树的最大深度

题目地址：https://leetcode.cn/problems/maximum-depth-of-binary-tree/description/?envType=study-plan-v2&envId=top-interview-150

递归找左子树和右子树的最大深度，就是二叉树的最大深度。

# 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 = rightclass Solution:def maxDepth(self, root: Optional[TreeNode]) -> int:if not root:return 0else:left_depth = self.maxDepth(root.left)right_depth = self.maxDepth(root.right)return max(left_depth, right_depth) + 1

9.2 【递归】相同的树

题目地址：https://leetcode.cn/problems/same-tree/description/?envType=study-plan-v2&envId=top-interview-150

左右子树相同的标志是左右子树都存在，且根节点的值相等，按照这一标准递归遍历树的左右子树。

# 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 = rightclass Solution:def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:if (not p and q) or (p and not q):return Falseelif not p and not q:return Trueelif p.val != q.val:return Falseelse:return self.isSameTree(p.left,q.left) and self.isSameTree(p.right,q.right)

9.3 【递归】翻转二叉树

题目地址：https://leetcode.cn/problems/invert-binary-tree/description/?envType=study-plan-v2&envId=top-interview-150

详见代码。

# 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 = rightclass Solution:def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:if not root:returnroot.left,root.right = self.invertTree(root.right),self.invertTree(root.left)return root

9.4 【递归】对称二叉树

题目地址：https://leetcode.cn/problems/symmetric-tree/description/?envType=study-plan-v2&envId=top-interview-150

详见代码。

# 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 isSymmetric(self, root: Optional[TreeNode]) -> bool:if not root:return Truereturn self.is_symmetric(root.left, root.right)def is_symmetric(self, left: TreeNode, right: TreeNode):if not left and not right:return Trueelif (not left or not right) or (left.val != right.val):return Falseelse:return self.is_symmetric(left.left, right.right) and self.is_symmetric(left.right, right.left)

9.5 【递归】从前序与中序遍历序列构造二叉树

题目地址：https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/description/?envType=study-plan-v2&envId=top-interview-150

详见代码，找到根节点的位置进行递归。

# 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 = rightclass Solution:def buildTree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:if not preorder or not inorder:returnroot = TreeNode(preorder[0])idx = inorder.index(preorder[0])root.left = self.buildTree(preorder[1:1+idx],inorder[:idx])root.right = self.buildTree(preorder[1+idx:],inorder[idx+1:])return root

9.6 【递归】从中序与后序遍历序列构造二叉树

题目地址：https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/description/?envType=study-plan-v2&envId=top-interview-150

详见代码，找到根节点的位置进行递归。

# 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 buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:if not inorder or not postorder:returnroot = TreeNode(postorder[-1])idx = inorder.index(postorder[-1])root.left = self.buildTree(inorder[:idx],postorder[:idx])root.right = self.buildTree(inorder[1+idx:],postorder[idx:-1])return root

9.7 【BFS】填充每个节点的下一个右侧节点指针 II

题目地址：https://leetcode.cn/problems/populating-next-right-pointers-in-each-node-ii/description/?envType=study-plan-v2&envId=top-interview-150

其实就是找出每一层的所有结点就行了，而每一层的左右节点就是下一层的结点，按照这个规则进行BFS。

"""
# Definition for a Node.
class Node:def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):self.val = valself.left = leftself.right = rightself.next = next
"""class Solution:def connect(self, root: 'Node') -> 'Node':if not root:return rootdef BFS(curLayer):nextLayer = []for node in curLayer:if node.left:nextLayer.append(node.left)if node.right:nextLayer.append(node.right)if len(nextLayer) > 1:for i in range(0,len(nextLayer)-1):nextLayer[i].next = nextLayer[i+1]if nextLayer:BFS(nextLayer)BFS([root])return root

9.8 【递归】二叉树展开为链表

题目地址：https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/description/?envType=study-plan-v2&envId=top-interview-150

二叉树的先序遍历的变形。

# 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 = rightclass Solution:def flatten(self, root: Optional[TreeNode]) -> None:"""Do not return anything, modify root in-place instead."""while root:if root.left:cur_left = root.leftwhile cur_left.right:cur_left = cur_left.rightcur_left.right = root.rightroot.right = root.leftroot.left = Noneroot = root.right

9.9 【DFS】路径总和

题目地址：https://leetcode.cn/problems/path-sum/description/?envType=study-plan-v2&envId=top-interview-150

详见代码。

# 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 = rightclass Solution:def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool:if not root:return Falseelif not root.left and not root.right and targetSum == root.val:return Trueelse:return self.hasPathSum(root.left,targetSum-root.val) or self.hasPathSum(root.right,targetSum-root.val)

9.10 【DFS】求根节点到叶节点数字之和

题目地址：https://leetcode.cn/problems/sum-root-to-leaf-numbers/description/?envType=study-plan-v2&envId=top-interview-150

DFS，分别从根节点计算每一条路径的数字，然后求和，这里要注意判断条件，当一个节点没有左节点和右节点时，说明这条路径到底了，这是要把目前的数字加入到 $an s$ 中。

# 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 = rightclass Solution:def sumNumbers(self, root: Optional[TreeNode]) -> int:ans = 0def sub_sum(root,cur_num):nonlocal ansif not root.left and not root.right:ans += cur_num*10 + root.valreturncur_num = cur_num*10 + root.valif root.left:sub_sum(root.left,cur_num)if root.right:sub_sum(root.right,cur_num)sub_sum(root,0)return ans

9.11 【BFS】【动态规划】二叉树中的最大路径和

题目地址：https://leetcode.cn/problems/binary-tree-maximum-path-sum/description/?envType=study-plan-v2&envId=top-interview-150

先找到每个节点下的最大路径和，再一步步从上往下遍历所有节点，如果遍历到某个节点的左子树或者右子树的最大和小于零，则这个子树可以不要，只需考虑另一个子树即可。

# 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 = rightclass Solution:def maxPathSum(self, root: Optional[TreeNode]) -> int:self.maxSum = -1001def BFS(root):if not root:return 0left_max = BFS(root.left)right_max = BFS(root.right)self.maxSum = max(self.maxSum, left_max + right_max + root.val)return max(0,max(left_max,right_max) + root.val)BFS(root)return self.maxSum

9.12 【BFS】二叉搜索树迭代器

题目地址：https://leetcode.cn/problems/binary-search-tree-iterator/description/?envType=study-plan-v2&envId=top-interview-150

先对二叉搜索树中的元素进行排序，然后按照题目要求构造函数即可。

# 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 = rightclass BSTIterator:def __init__(self, root: Optional[TreeNode]):self.root = rootself.num = []self.idx = 0self.len = 0def BFS(root):if not root:returnself.num.append(root.val)self.len += 1BFS(root.left)BFS(root.right)BFS(root)self.num.sort()def next(self) -> int:if self.idx < self.len:self.idx += 1return self.num[self.idx-1]else:return -1def hasNext(self) -> bool:if self.idx < self.len:return Trueelse:return False# Your BSTIterator object will be instantiated and called as such:
# obj = BSTIterator(root)
# param_1 = obj.next()
# param_2 = obj.hasNext()

9.13 【BFS】完全二叉树的节点个数

题目地址：https://leetcode.cn/problems/count-complete-tree-nodes/description/?envType=study-plan-v2&envId=top-interview-150

详见代码。

# 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 = rightclass Solution:def countNodes(self, root: Optional[TreeNode]) -> int:self.ans = 0def BFS(root):if not root:returnself.ans += 1BFS(root.left)BFS(root.right)BFS(root)return self.ans

9.14 【递归】二叉树的最近公共祖先

题目地址：https://leetcode.cn/problems/lowest-common-ancestor-of-a-binary-tree/description/?envType=study-plan-v2&envId=top-interview-150

如果 $p$ 和 $q$ 位于二叉树的同一侧，那么最近公共祖先要么是 $p$ 要么是 $q$ ，如果位于二叉树的两侧，那么最近公共祖先则是两者最深的那个 $roo t$ 。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = Noneclass Solution:def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':if not root or root == p or root == q:return rootleft = self.lowestCommonAncestor(root.left,p,q)right = self.lowestCommonAncestor(root.right,p,q)if not left:return rightif not right:return leftreturn root

10 二叉树层次遍历

10.1 【DFS】二叉树的右视图

题目地址：https://leetcode.cn/problems/binary-tree-right-side-view/description/?envType=study-plan-v2&envId=top-interview-150

深度优先遍历，每次都将二叉树每一层最右边的元素加入数组，同时设置 $d e pt h$ 来限制数组大小，保证数组中只有最右边的元素。

# 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 = rightclass Solution:def rightSideView(self, root: Optional[TreeNode]) -> List[int]:self.ans = []def DFS(root, depth):if not root:returnif len(self.ans) < depth:self.ans.append(root.val)DFS(root.right, depth+1)DFS(root.left, depth+1)DFS(root, 1)return self.ans

10.2 【BFS】二叉树的层平均值

题目地址：https://leetcode.cn/problems/average-of-levels-in-binary-tree/description/?envType=study-plan-v2&envId=top-interview-150

详见代码。

# 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 = rightclass Solution:def averageOfLevels(self, root: Optional[TreeNode]) -> List[float]:ans = []node_list =  [root]if not root:return answhile node_list:cur_val = []nxt_list = []for node in node_list:cur_val.append(node.val)if node.left:nxt_list.append(node.left)if node.right:nxt_list.append(node.right)ans.append(sum(cur_val)/len(cur_val))node_list = nxt_listreturn ans

10.3 【BFS】二叉树的层序遍历

题目地址：https://leetcode.cn/problems/binary-tree-level-order-traversal/description/?envType=study-plan-v2&envId=top-interview-150

详见代码。

# 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 = rightclass Solution:def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:ans = []cur_node = [root]if not root:return answhile cur_node:cur_val = []nxt_node = []for node in cur_node:cur_val.append(node.val)if node.left:nxt_node.append(node.left)if node.right:nxt_node.append(node.right)ans.append(cur_val)cur_node = nxt_nodereturn ans

10.4 【BFS】二叉树的锯齿形层序遍历

题目地址：https://leetcode.cn/problems/binary-tree-zigzag-level-order-traversal/description/?envType=study-plan-v2&envId=top-interview-150

详见代码。

# 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 = rightclass Solution:def zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:ans = []cur_node = [root]flag = Trueif not root:return answhile cur_node:cur_val = []nxt_node = []for node in cur_node:cur_val.append(node.val)if node.left:nxt_node.append(node.left)if node.right:nxt_node.append(node.right)if flag:ans.append(cur_val)else:ans.append(cur_val[::-1])flag = not flagcur_node = nxt_nodereturn ans

11 二叉搜索树

11.1 【BFS】二叉搜索树的最小绝对差

题目地址：https://leetcode.cn/problems/minimum-absolute-difference-in-bst/description/?envType=study-plan-v2&envId=top-interview-150

先遍历所有节点，然后排序，找出相邻元素差值最小的。

# 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 = rightclass Solution:def getMinimumDifference(self, root: Optional[TreeNode]) -> int:self.num = []def BFS(root):if not root:returnself.num.append(root.val)BFS(root.left)BFS(root.right)BFS(root)self.num.sort()ans = 100001for i in range(len(self.num)-1):ans = min(ans,self.num[i+1]-self.num[i])return ans

11.2 【BFS】二叉搜索树中第K小的元素

题目地址：https://leetcode.cn/problems/kth-smallest-element-in-a-bst/description/?envType=study-plan-v2&envId=top-interview-150

对于二叉搜索树而言，中序遍历就是其从小到大的排序结果，记录访问到的第 $k$ 个元素即可。

# 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 = rightclass Solution:def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:self.k = kself.ans = 0def BFS(root):if not root:returnBFS(root.left)self.k -= 1if self.k == 0:self.ans = root.valreturnBFS(root.right)BFS(root)return self.ans

11.3 【BFS】【递归】验证二叉搜索树

题目地址：https://leetcode.cn/problems/validate-binary-search-tree/description/?envType=study-plan-v2&envId=top-interview-150

递归实现，首先要明白二叉搜索树的判定条件，一是左子树<root<右子树，二是下面的左子树的值要比当前结点小，而右子树的值要比当前结点大，也就是不能只在一个小的子树上满足这个条件，往上回溯几代之后这个条件要依然满足才行。

# 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 = rightclass Solution:def isValidBST(self, root: Optional[TreeNode]) -> bool:def is_Valid_BST(root, min_node, max_node):if not root:return Trueelse:if min_node and root.val <= min_node.val:return Falseif max_node and root.val >= max_node.val:return Falsereturn is_Valid_BST(root.left, min_node, root) and is_Valid_BST(root.right, root, max_node)return is_Valid_BST(root, None, None)