Question

Formatted question description: https://leetcode.ca/all/117.html

Given a binary tree

struct Node {
  int val;
  Node *left;
  Node *right;
  Node *next;
}

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Example 1:

Input: root = [1,2,3,4,5,null,7]
Output: [1,#,2,3,#,4,5,7,#]
Explanation: Given the above binary tree (Figure A), your function should populate each next pointer to point to its next right node, just like in Figure B. The serialized output is in level order as connected by the next pointers, with '#' signifying the end of each level.

Example 2:

Input: root = []
Output: []

Constraints:

The number of nodes in the tree is in the range [0, 6000].
-100 <= Node.val <= 100

Follow-up:

You may only use constant extra space.
The recursive approach is fine. You may assume implicit stack space does not count as extra space for this problem.

Algorithm

BFS

Since the subtree may be incomplete, it is necessary to scan the nodes in the same layer of the parent node in parallel to find their left and right child nodes.

DFS

Same as in 116 - Populating Next Right Pointers in Each Node.

Code

public class Populating_Next_Right_Pointers_in_Each_Node_II {

    public static void main(String[] args) {
        Populating_Next_Right_Pointers_in_Each_Node_II out = new Populating_Next_Right_Pointers_in_Each_Node_II();

        Solution_fancy s = out.new Solution_fancy();

        TreeLinkNode root = new TreeLinkNode(1);
        root.left = new TreeLinkNode(2);
        root.right = new TreeLinkNode(3);
        root.left.left = new TreeLinkNode(4);
        root.left.right = new TreeLinkNode(5);
        root.right.left = new TreeLinkNode(6);
        root.right.right = new TreeLinkNode(7);

        s.connect(root);
    }

    /**
     * Definition for binary tree with next pointer.
     * public class TreeLinkNode {
     * int val;
     * TreeLinkNode left, right, next;
     * TreeLinkNode(int x) { val = x; }
     * }
     */

    public class Solution {
        public void connect(TreeLinkNode root) {

            if (root == null) {
                return;
            }

            Queue<TreeLinkNode> q = new LinkedList<>();

            TreeLinkNode levelEndMarker = new TreeLinkNode(0);
            TreeLinkNode prev = new TreeLinkNode(0);

            q.offer(root);
            q.offer(levelEndMarker);

            while (!q.isEmpty()) {
                TreeLinkNode current = q.poll();

                if (current == levelEndMarker) {
                    // push marker again as next level end
                    if (!q.isEmpty()) { // @note: missed final check again...
                        q.offer(levelEndMarker);
                    }

                    // first node of each level, can not be last of upper level
                    prev = levelEndMarker; // can be any node which is not in this tree
                    continue;
                }

                if (current.left != null) {
                    q.offer(current.left);
                }
                if (current.right != null) {
                    q.offer(current.right);
                }

                prev.next = current;
                prev = current;
            }
        }
    }

    class Solution_fancy {
        public TreeLinkNode connect(TreeLinkNode root) {
            TreeLinkNode dummyHead = new TreeLinkNode(0); // this head will always point to the first element in the current layer we are searching
            TreeLinkNode pre = dummyHead; // this 'pre' will be the "current node" that builds every single layer
            TreeLinkNode real_root = root; // just for return statement

            while (root != null) {
                if (root.left != null) {
                    pre.next = root.left; // @note: here pre is same as dummyHead, pointing to 1st node of this level. this is before pre is updated to be other nodes
                    pre = pre.next;
                }
                if (root.right != null) {
                    pre.next = root.right;
                    pre = pre.next;
                }
                root = root.next;
                if (root == null) { // reach the end of current layer
                    pre = dummyHead; // shift pre back to the beginning, get ready to point to the first element in next layer
                    root = dummyHead.next;
                    ;//root comes down one level below to the first available non null node
                    dummyHead.next = null;//reset dummyhead back to default null, so that later it will point to next level's first node
                }
            }
            return real_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;
    }
}

// OJ: https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
// Time: O(N)
// Space: O(H)
class Solution {
public:
    Node* connect(Node* root) {
        if (!root) return nullptr;
        Node head, *tail = &head;
        for (auto node = root; node; node = node->next) {
            if (node->left) {
                tail->next = node->left;
                tail = tail->next;
            }
            if (node->right) {
                tail->next = node->right;
                tail = tail->next;
            }
        }
        connect(head.next);
        return root;
    }
};

"""
# 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_node, next_node
            if curr is None:
                return
            if next_node is None: # next level's first node
                next_node = curr
            # "if next_node is None" logic can be replaced by:
            #   next_node = next_node or curr
            if prev_node:
                prev_node.next = curr
            prev_node = curr

        node = root
        while node:
            prev_node = next_node = None
            while node: # process for every level
                modify(node.left)
                modify(node.right)
                node = node.next
            node = next_node
        return root



# use dummyHead.next to find each level's first node
class Solution:
    def connect(self, root: 'Node') -> 'Node':
        dummyHead = Node(0)
        pre = dummyHead
        real_root = root

        while root:
            if root.left:
                # @note: here pre is same as dummyHead, pointing to 1st node of this level. this is before pre is updated to be other nodes
                pre.next = root.left
                pre = pre.next
            if root.right:
                pre.next = root.right
                pre = pre.next
            root = root.next
            if not root: # reach the end of current layer
                pre = dummyHead # shift pre back to the beginning, get ready to point to the first element in next layer
                root = dummyHead.next # root comes down one level below to the first available non null node
                dummyHead.next = None # reset dummyhead back to default null, so that later it will point to next level's first node

        return real_root

###############


# Definition for binary tree with next pointer.
# class TreeLinkNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
  # @param root, a tree link node
  # @return nothing
  def connect(self, root):
    p = root
    pre = None
    head = None
    while p:
      if p.left:
        if pre:
          pre.next = p.left
        pre = p.left
      if p.right:
        if pre:
          pre.next = p.right
        pre = p.right
      if not head:
        head = p.left or p.right
      if p.next:
        p = p.next
      else:
        p = head
        head = None
        pre = None

/**
 * 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 {
    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;
            right && queue.push(right);
            left && queue.push(left);
        }
    }
    return root;
}

117 - Populating Next Right Pointers in Each Node II

Question

Algorithm

BFS

DFS

Code

All Problems

All Solutions

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Algorithm

BFS

DFS

Code

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