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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;
    }
    
    

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