递归解决问题思路

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/*
先遍历节点，记录所有index, map的key是when的index, 值是这个when对应的要执行的代码对应的index
map = {
 0: [1, 2, 3],
 4: [5, 6],
 7: [8],
 9: [10, 11]
}
*/

var map = {};
var whenIndexList = [] // 记录所有when的index
var curWhenIndex = 0;
for (var i = 0; i < statements.length; i++) {
 map[curWhenIndex] = [];
 如果没有匹配到when, map[curWhenIndex].push(i);
 如果匹配到when，curWhenIndex = i;

}

// 遍历map来拼接
var whens = Object.keys(map);
var str = '';

for (var i = 0; i < whens.length; i++) {
 var notWhen = map[whens[i]];
 for (var j = 0; j < notWhen.length; j++) {
  str += statements[notWhen[j]];
 }
}