题目:
难度:Middle
相关话题:树
、深度优先搜索
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}
解释:给定二叉树如图 A 所示,你的函数应该填充它的每个 next 指针,以指向其下一个右侧节点,如图 B 所示。
提示:
思路:
这道题要求不能使用额外空间 或者 使用递归。
递归的思想就是,对于左支leftNode
和右支rightNode
,先分别将他们内部左支和右支的next
连接, 再将左支leftNode
内最右的子支和右支rightNode
内最左的子支的next
连接。
/**
* @来源: Javascript中文网 - 前端进阶资源教程 https://www.javascriptc.com/
* @介绍:一个致力于帮助开发者用代码改变世界为使命的平台,每天都可以在这里找到技术世界的头条内容
* // Definition for a Node.
* function Node(val,left,right,next) {
* this.val = val;
* this.left = left;
* this.right = right;
* this.next = next;
* };
*/
/**
* @param {Node} root
* @return {Node}
*/
var connect = function(root) {
if(!root)return null
function getLeft(root){
return root.left || root.right
}
function getRight(root){
return root.right || root.left
}
function con(root1,root2){
if(!root1 || !root2)return
root1.next=root2
con(root1.left,root1.right)
con(root2.left,root2.right)
con(getRight(root1),getLeft(root2))
}
con(root.left,root.right)
return root
};