题解 | #重建二叉树#

本来的思路是——关注节点关系的变化，构思是 p1->p3, p2->p1, p3->p2......, 所以写出了下面的代码：

function ReverseList(pHead) {
  if (!pHead) {
    return null;
  }

  let curNode = pHead;
  let next = pHead.next;
  while (next) {
    pHead.next = next.next;
    next.next = curNode;
    curNode = next;
    next = pHead.next;
  }

  return curNode;
}

殊不知节点关系已经存在两个声明的变量里了，三节点法是利用率更高的

function ReverseList(pHead) {
    // write code here
    let pre = null;
    let next = null;
    while(pHead !== null) {
        next = pHead.next;
        pHead.next = pre;
        pre = pHead;
        pHead = next;
    }
    return pre;
}