二叉树层序遍历登场！

• 102. 二叉树的层序遍历

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
 var q []*TreeNode
var head int
var tail int

func init() {
	q = make([]*TreeNode, 10010)
	head = -1
	tail = -1
}
func empty() bool {
	return head == tail
}
func push(x *TreeNode) {
	tail++
	q[tail] = x
}
func pop() *TreeNode {
	head++
	return q[head]
}
func size() int {
	return tail - head
}
func levelOrder(root *TreeNode) [][]int {
	var res [][]int
	
	push(root)
	for !empty() {
		qSize := size() // 细节
		var temp []int
		for i := 0; i < qSize; i++ {
			node := pop()
			if node == nil {
				continue
			}
			// watch
			temp = append(temp, node.Val)
			push(node.Left)
			push(node.Right)
		}
		if len(temp) > 0 {
			res = append(res, temp)
		}
	}
	return res
}

• 107. 二叉树的层序遍历 II

	slices.Reverse(res)
	return res

226.翻转二叉树

func invertTree(root *TreeNode) *TreeNode {
    if root == nil {
        return nil
    }
    left := invertTree(root.Left)
    right :=invertTree(root.Right)
    root.Left = right
    root.Right = left
    return root
}

101. 对称二叉树

• 后序遍历，但是一边要反后序遍历（一个左右，一个右左）

func check (left, right *TreeNode) bool{
    if left == nil && right == nil {
        return true
    } 
    if (left == nil && right != nil) || (left != nil && right == nil) {
        return false
    }
    if left.Val != right.Val {
        return false
    }
    return check(left.Left, right.Right) && check(left.Right, right.Left)
 }
func isSymmetric(root *TreeNode) bool {
    return check(root.Left, root.Right)
}