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116. 填充每个节点的下一个右侧节点指针

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

方法一：BFS

使用队列进行层序遍历，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: "Optional[Node]") -> "Optional[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 == null || root.left == null) {
        return root;
    }
    const { left, right, next } = root;
    left.next = right;
    if (next != null) {
        right.next = next.left;
    }
    connect(left);
    connect(right);
    return root;
}

方法二：DFS

使用递归进行前序遍历，每次遍历到一个节点时，将其左右子节点按顺序连接起来。

具体地，我们设计一个函数 $dfs(left, right)$，表示将 $left$ 节点的 $next$ 指针指向 $right$ 节点。在函数中，我们首先判断 $left$ 和 $right$ 是否为空，若都不为空，则将 $left.next$ 指向 $right$，然后递归地调用 $dfs(left.left, left.right)$, $dfs(left.right, right.left)$, $dfs(right.left, right.right)$。

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: 'Optional[Node]') -> 'Optional[Node]':
        def dfs(left, right):
            if left is None or right is None:
                return
            left.next = right
            dfs(left.left, left.right)
            dfs(left.right, right.left)
            dfs(right.left, right.right)

        if root:
            dfs(root.left, root.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) {
            dfs(root.left, root.right);
        }
        return root;
    }

    private void dfs(Node left, Node right) {
        if (left == null || right == null) {
            return;
        }
        left.next = right;
        dfs(left.left, left.right);
        dfs(left.right, right.left);
        dfs(right.left, right.right);
    }
}

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) {
        function<void(Node*, Node*)> dfs = [&](Node* left, Node* right) {
            if (!left || !right) {
                return;
            }
            left->next = right;
            dfs(left->left, left->right);
            dfs(left->right, right->left);
            dfs(right->left, right->right);
        };
        if (root) {
            dfs(root->left, root->right);
        }
        return root;
    }
};

Go

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

func connect(root *Node) *Node {
	var dfs func(*Node, *Node)
	dfs = func(left, right *Node) {
		if left == nil || right == nil {
			return
		}
		left.Next = right
		dfs(left.Left, left.Right)
		dfs(left.Right, right.Left)
		dfs(right.Left, right.Right)
	}
	if root != nil {
		dfs(root.Left, root.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 == null) {
        return root;
    }
    const queue = [root];
    while (queue.length !== 0) {
        const n = queue.length;
        let pre = null;
        for (let i = 0; i < n; i++) {
            const node = queue.shift();
            node.next = pre;
            pre = node;
            const { left, right } = node;
            left && queue.push(right, left);
        }
    }
    return root;
}

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116. 填充每个节点的下一个右侧节点指针

题目描述

解法

方法一：BFS

Python3

Java

C++

Go

TypeScript

方法二：DFS

Python3

Java

C++

Go

TypeScript

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116. 填充每个节点的下一个右侧节点指针

题目描述

解法

方法一：BFS

Python3

Java

C++

Go

TypeScript

方法二：DFS

Python3

Java

C++

Go

TypeScript