Leetcode 102. Sequence traversal of binary trees

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Given a binary tree, return the values of nodes traversed in order. (That is, layer by layer, all nodes are accessed from left to right).

That is, return a two-dimensional array.

Implementation (iteration, commit pass)

import "container/list"

/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */
func levelOrder(root *TreeNode)[] []int {
    if root == nil {
        return nil
    }
    var (
        res [][]int
        q = NewQueue()
    )
    q.Push(root)
    for! q.IsEmpty() { c := q.Len() level :=make([]int.0, c)
        for i := 0; i < c; i++ {
            node := q.Pop().(*TreeNode)
            level = append(level, node.Val)
            ifnode.Left ! =nil {
                q.Push(node.Left)
            }
            ifnode.Right ! =nil {
                q.Push(node.Right)
            }
        }
        res = append(res, level)
    }
    return res
}

type Queue struct {
	l *list.List
}

func NewQueue(a) *Queue {
	return &Queue{
		l: list.New(),
	}
}

func (o *Queue) Push(v interface{}) {
	o.l.PushBack(v)
}

func (o *Queue) Pop(a) interface{} {
	if o.IsEmpty() {
		panic("Queue is empty")
	}
	e := o.l.Front()
	o.l.Remove(e)
	return e.Value
}

func (o *Queue) Front(a) interface{} {
	if o.IsEmpty() {
		panic("Queue is empty")}return o.l.Front().Value
}

func (o *Queue) Back(a) interface{} {
	if o.IsEmpty() {
		panic("Queue is empty")}return o.l.Back().Value
}

func (o *Queue) IsEmpty(a) bool {
	return o.l.Len() == 0
}

func (o *Queue) Len(a) int {
	return o.l.Len()
}
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