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]]

# 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]]

# 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

# 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