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117. 填充每个节点的下一个右侧节点指针 II

题目描述

给定一个二叉树：

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,null,7]
输出：[1,#,2,3,#,4,5,7,#]
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。序列化输出按层序遍历顺序（由 next 指针连接），'#' 表示每层的末尾。

示例 2：

输入：root = []
输出：[]

提示：

树中的节点数在范围 [0, 6000] 内
-100 <= Node.val <= 100

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，本题中递归程序的隐式栈空间不计入额外空间复杂度。

解法

方法一：BFS

我们使用队列 $q$ 进行层序遍历，每次遍历一层时，将当前层的节点按顺序连接起来。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树的节点个数。

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:
            return root
        q = deque([root])
        while q:
            p = None
            for _ in range(len(q)):
                node = q.popleft()
                if p:
                    p.next = node
                p = node
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
        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) {
            return root;
        }
        Deque<Node> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            Node p = null;
            for (int n = q.size(); n > 0; --n) {
                Node node = q.poll();
                if (p != null) {
                    p.next = node;
                }
                p = node;
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
        }
        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) {
            return root;
        }
        queue<Node*> q{{root}};
        while (!q.empty()) {
            Node* p = nullptr;
            for (int n = q.size(); n; --n) {
                Node* node = q.front();
                q.pop();
                if (p) {
                    p->next = node;
                }
                p = node;
                if (node->left) {
                    q.push(node->left);
                }
                if (node->right) {
                    q.push(node->right);
                }
            }
        }
        return root;
    }
};

Go

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Left *Node
 *     Right *Node
 *     Next *Node
 * }
 */

func connect(root *Node) *Node {
	if root == nil {
		return root
	}
	q := []*Node{root}
	for len(q) > 0 {
		var p *Node
		for n := len(q); n > 0; n-- {
			node := q[0]
			q = q[1:]
			if p != nil {
				p.Next = node
			}
			p = node
			if node.Left != nil {
				q = append(q, node.Left)
			}
			if node.Right != nil {
				q = append(q, node.Right)
			}
		}
	}
	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) {
        return null;
    }
    const q: Node[] = [root];
    while (q.length) {
        const nq: Node[] = [];
        let p: Node | null = null;
        for (const node of q) {
            if (p) {
                p.next = node;
            }
            p = node;
            const { left, right } = node;
            left && nq.push(left);
            right && nq.push(right);
        }
        q.splice(0, q.length, ...nq);
    }
    return root;
}

C#

/*
// Definition for a Node.
public 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;
    }
}
*/

public class Solution {
    public Node Connect(Node root) {
        if (root == null) {
            return null;
        }
        var q = new Queue<Node>();
        q.Enqueue(root);
        while (q.Count > 0) {
            Node p = null;
            for (int i = q.Count; i > 0; --i) {
                var node = q.Dequeue();
                if (p != null) {
                    p.next = node;
                }
                p = node;
                if (node.left != null) {
                    q.Enqueue(node.left);
                }
                if (node.right != null) {
                    q.Enqueue(node.right);
                }
            }
        }
        return root;
    }
}

方法二：空间优化

方法一的空间复杂度较高，因为需要使用队列存储每一层的节点。我们可以使用常数空间来实现。

定义两个指针 $prev$ 和 $next$，分别指向下一层的前一个节点和第一个节点。遍历当前层的节点时，把下一层的节点串起来，同时找到下一层的第一个节点。当前层遍历完后，把下一层的第一个节点 $next$ 赋值给 $node$，继续遍历。

时间复杂度 $O(n)$，其中 $n$ 为二叉树的节点个数。空间复杂度 $O(1)$。

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':
        def modify(curr):
            nonlocal prev, next
            if curr is None:
                return
            next = next or curr
            if prev:
                prev.next = curr
            prev = curr

        node = root
        while node:
            prev = next = None
            while node:
                modify(node.left)
                modify(node.right)
                node = node.next
            node = next
        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 {
    private Node prev, next;

    public Node connect(Node root) {
        Node node = root;
        while (node != null) {
            prev = null;
            next = null;
            while (node != null) {
                modify(node.left);
                modify(node.right);
                node = node.next;
            }
            node = next;
        }
        return root;
    }

    private void modify(Node curr) {
        if (curr == null) {
            return;
        }
        if (next == null) {
            next = curr;
        }
        if (prev != null) {
            prev.next = curr;
        }
        prev = curr;
    }
}

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) {
        Node* node = root;
        Node* prev = nullptr;
        Node* next = nullptr;
        auto modify = [&](Node* curr) {
            if (!curr) {
                return;
            }
            if (!next) {
                next = curr;
            }
            if (prev) {
                prev->next = curr;
            }
            prev = curr;
        };
        while (node) {
            prev = next = nullptr;
            while (node) {
                modify(node->left);
                modify(node->right);
                node = node->next;
            }
            node = next;
        }
        return root;
    }
};

Go

/**
 * Definition for a Node.
 * type Node struct {
 *     Val int
 *     Left *Node
 *     Right *Node
 *     Next *Node
 * }
 */

func connect(root *Node) *Node {
	node := root
	var prev, next *Node
	modify := func(curr *Node) {
		if curr == nil {
			return
		}
		if next == nil {
			next = curr
		}
		if prev != nil {
			prev.Next = curr
		}
		prev = curr
	}
	for node != nil {
		prev, next = nil, nil
		for node != nil {
			modify(node.Left)
			modify(node.Right)
			node = node.Next
		}
		node = next
	}
	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 {
    const modify = (curr: Node | null): void => {
        if (!curr) {
            return;
        }
        next = next || curr;
        if (prev) {
            prev.next = curr;
        }
        prev = curr;
    };
    let node = root;
    let [prev, next] = [null, null];
    while (node) {
        while (node) {
            modify(node.left);
            modify(node.right);
            node = node.next;
        }
        node = next;
        [prev, next] = [null, null];
    }
    return root;
}

C#

/*
// Definition for a Node.
public 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;
    }
}
*/

public class Solution {
    private Node prev, next;

    public Node Connect(Node root) {
        Node node = root;
        while (node != null) {
            prev = null;
            next = null;
            while (node != null) {
                modify(node.left);
                modify(node.right);
                node = node.next;
            }
            node = next;
        }
        return root;
    }

    private void modify(Node curr) {
        if (curr == null) {
            return;
        }
        if (next == null) {
            next = curr;
        }
        if (prev != null) {
            prev.next = curr;
        }
        prev = curr;
    }
}

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117. 填充每个节点的下一个右侧节点指针 II

题目描述

解法

方法一：BFS

Python3

Java

C++

Go

TypeScript

C#

方法二：空间优化

Python3

Java

C++

Go

TypeScript

C#

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README.md

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117. 填充每个节点的下一个右侧节点指针 II

题目描述

解法

方法一：BFS

Python3

Java

C++

Go

TypeScript

C#

方法二：空间优化

Python3

Java

C++

Go

TypeScript

C#