103. 二叉树的锯齿形层序遍历

原题

https://leetcode.cn/problems/binary-tree-zigzag-level-order-traversal/description/

思路

广度优先搜索

复杂度

时间：O(n)
空间：O(n)

Python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 zigzagLevelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        if not root:
            return []
        ascending = True 
        ans = []
        q = [root]
        while q:
            level = []
            for i in range(len(q)):
                node = q.pop(0)
                level.append(node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            if ascending:
                ans.append(level)
            else:
                ans.append(level[::-1])
            ascending = not ascending
        return ans
       

Go代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func zigzagLevelOrder(root *TreeNode) [][]int {
	var ans [][]int
	if root == nil {
		return ans
	}
	// 是否反序
	reverse := false
	q := []*TreeNode{root}
	for len(q) > 0 {
		n := len(q)
		// 节点值的层序遍历
		vals := make([]int, n)
		for i := 0; i < n; i++ {
			node := q[0]
			q = q[1:]
			if reverse {
				// 倒序
				vals[n-1-i] = node.Val
			} else {
				vals[i] = node.Val
			}

			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
		ans = append(ans, vals)
		reverse = !reverse
	}
	return ans
}