题目描述
给定一个完美二叉树,其所有叶子节点都在同一层,每个父节点都有两个子节点。二叉树定义如下:
struct Node {
int val;
Node *left;
Node *right;
Node *next;
}
填充它的每个 next 指针,让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点,则将 next 指针设置为 NULL。
初始状态下,所有 next 指针都被设置为 NULL。
示例:
输入:{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
输出:{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}
思路
1.第一层连接第二层,第二层连接第三层,依次类推,迭代到最后。
2.使用cur标识本层正在移动的结点,pre标识cur结点的前一个结点,start标识本层第一个结点。
3.具体迭代过程,可以参考下图
Java代码实现
public Node connect(Node root) {
if(root == null)
return null;
//当前遍历的前一个节点
Node pre = root;
//当前遍历的节点
Node cur = null;
//每层开始的节点
Node start = pre;
while(start.left != null){
//如果本层走到了NULL,就代表需要进行下一层的处理
if(cur == null){
//连接上最后的两个节点
pre.left.next = pre.right;
//初始化pre节点
pre = start.left;
//初始化cur节点
cur = start.right;
//初始化start节点
start = pre;
}
//依次连接下层节点
else{
pre.left.next = pre.right;
pre.right.next = cur.left;
pre = pre.next;
cur = cur.next;
}
}
return root;
}
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