二叉树的层序遍历

2026-02-04

1. 二叉树的层序遍历

(102）给你二叉树的根节点 root ，返回其节点值的 层序遍历 。（即逐层地，从左到右访问所有节点）。

输入：root = [3,9,20,null,null,15,7]
输出：[[3],[9,20],[15,7]]
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```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 levelOrder1(self, root: Optional[TreeNode]) -> List[List[int]]:
       if root is None:
           return []
       cur = [root]
       res = []
       while cur:
           child = []
           res_ = []
           for node in cur:
               res_.append(node.val)
               if node.left:
                   child.append(node.left)
               if node.right:
                   child.append(node.right)
           res.append(res_)
           cur = child
       return res

   def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
       if root is None:
           return []
       cur = deque([root])
       res = []
       while cur:
           # child = []
           res_ = []
           for _ in range(len(cur)):
               node = cur.popleft()
               res_.append(node.val)
               if node.left:
                   cur.append(node.left)
               if node.right:
                   cur.append(node.right)
           res.append(res_)
           # cur = child
       return res

2. 二叉树的锯齿形层序遍历

给你二叉树的根节点 root ，返回其节点值的 锯齿形层序遍历 。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。

输入：root = [3,9,20,null,null,15,7]
输出：[[3],[20,9],[15,7]]
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```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 zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
       if root is None:
           return []
       cur = deque([root])
       res = []
       flag = True
       while(cur) :
           res_ = []
           for _ in range(len(cur)):
               node = cur.popleft()
               res_.append(node.val)
               if node.left:
                   cur.append(node.left)
               if node.right:
                   cur.append(node.right)
           res.append(res_ if flag else res_[::-1])
           flag = not flag
       return res

3. 找树左下角的值

（513）给定一个二叉树的 根节点 root，请找出该二叉树的 最底层最左边 节点的值。

假设二叉树中至少有一个节点。

输入: root = [2,1,3]
输出: 1
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```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 findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
       q = deque([root])
       while q:
           node = q.popleft()
           if node.right:
               q.append(node.right)
           if node.left:
               q.append(node.left)
       return node.val