116. 填充每个节点的下一个右侧节点指针

English Version

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 1：

输入：root = [1,2,3,4,5,6,7]
输出：[1,#,2,3,#,4,5,6,7,#]
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。序列化的输出按层序遍历排列，同一层节点由 next 指针连接，'#' 标志着每一层的结束。

示例 2:

输入：root = []
输出：[]

 

提示：

• 树中节点的数量在 [0, 212 - 1] 范围内
• -1000 <= node.val <= 1000

 

进阶：

• 你只能使用常量级额外空间。
• 使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。

解法

“BFS 层次遍历”实现。

Python3

"""
# Definition for a Node.
class Node:
    def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None):
        self.val = val
        self.left = left
        self.right = right
        self.next = next
"""

class Solution:
    def connect(self, root: 'Node') -> 'Node':
        if root is None or (root.left is None and root.right is None):
            return root
        q = deque([root])
        while q:
            size = len(q)
            cur = None
            for _ in range(size):
                node = q.popleft()
                if node.right:
                    q.append(node.right)
                if node.left:
                    q.append(node.left)
                node.next = cur
                cur = node
        return root

Java

/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}

    public Node(int _val) {
        val = _val;
    }

    public Node(int _val, Node _left, Node _right, Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/

class Solution {
    public Node connect(Node root) {
        if (root == null || (root.left == null && root.right == null)) {
            return root;
        }
        Deque<Node> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            Node cur = null;
            for (int i = 0, n = q.size(); i < n; ++i) {
                Node node = q.pollFirst();
                if (node.right != null) {
                    q.offer(node.right);
                }
                if (node.left != null) {
                    q.offer(node.left);
                }
                node.next = cur;
                cur = node;
            }
        }
        return root;
    }
}

C++

/*
// Definition for a Node.
class Node {
public:
    int val;
    Node* left;
    Node* right;
    Node* next;

    Node() : val(0), left(NULL), right(NULL), next(NULL) {}

    Node(int _val) : val(_val), left(NULL), right(NULL), next(NULL) {}

    Node(int _val, Node* _left, Node* _right, Node* _next)
        : val(_val), left(_left), right(_right), next(_next) {}
};
*/

class Solution {
public:
    Node* connect(Node* root) {
        if (!root || (!root->left && !root->right)) {
            return root;
        }
        queue<Node*> q;
        q.push(root);
        while (!q.empty()) {
            Node* cur = nullptr;
            for (int i = 0, n = q.size(); i < n; ++i) {
                Node* node = q.front();
                q.pop();
                if (node->right) {
                    q.push(node->right);
                }
                if (node->left) {
                    q.push(node->left);
                }
                node->next = cur;
                cur = node;
            }
        }
        return root;
    }
};

TypeScript

/**
 * Definition for Node.
 * class Node {
 *     val: number
 *     left: Node | null
 *     right: Node | null
 *     next: Node | null
 *     constructor(val?: number, left?: Node, right?: Node, next?: Node) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function connect(root: Node | null): Node | null {
    if (root == null) {
        return root;
    }
    const { left, right } = root;
    if (left != null) {
        left.next = right;
    }
    if (right != null && root.next != null) {
        root.right.next = root.next.left;
    }
    connect(root.left);
    connect(root.right);
    return root;
}

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