102. 二叉树的层序遍历

原题

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

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 levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:
        ans = []
        if not root:
            return ans 
        curr_level = [root]
        while curr_level:
            l = []
            next_level = []
            for node in curr_level:
                l.append(node.val)
                if node.left:
                    next_level.append(node.left)
                if node.right:
                    next_level.append(node.right)
            ans.append(l)
            curr_level = next_level

        return ans 
       

Go代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func levelOrder(root *TreeNode) [][]int {
	var ans [][]int
	if root == nil {
		return ans
	}
	curr_level := []*TreeNode{root}
	for len(curr_level) > 0 {
		var l []int
		next_level := []*TreeNode{}
		for _, node := range curr_level {
			l = append(l, node.Val)
			if node.Left != nil {
				next_level = append(next_level, node.Left)
			}
			if node.Right != nil {
				next_level = append(next_level, node.Right)
			}
		}
		ans = append(ans, l)
		curr_level = next_level
	}
	return ans
}