写一个 traverse 函数,输出所有页面宽度和高度大于 50 像素的节点

function traverse(){
	var arr = [];
	var elements = [];
	if(document.all){
		elements = doucment.all;
	}else{
		elements = document.getElementsByTagName("*");
	}
	for(var i=0;i<elements.length;i++){
		var ele = elements[i];
		var width = parseFloat(ele.style.width) || ele.offsetWidth;
		var height = parseFloat(ele.style.height) || ele.offsetHeight;
		if(width > 50 && height >50){
			arr.push(elements[i].tagName);
		}
	}
	return arr;
	window.onload = function(){
		console.log(traverse());
		console.log("a");
		console.log('a');
	}
}	


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转载自blog.csdn.net/qq_21325977/article/details/80763775