for i := 0; i < len(levelOrder) / 2; i++ { levelOrder[i], levelOrder[len(levelOrder) - 1 - i] = levelOrder[len(levelOrder) - 1 - i], levelOrder[i] }
把等号右边的两项分别赋值给等号左边的两项a,b=c,d等价于a=cb=d