Question

117. 填充每个节点的下一个右侧节点指针 II

English Version

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。

"""
# 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

/*
// 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;
  }
}

/*
// 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

/*
// 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 $O(1)$。

"""
# 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

/*
// 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;
  }
}

/*
// 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

/*
// 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;
  }
}