Question

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

117. Populating Next Right Pointers in Each Node II

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:
    You may only use constant extra space.

For example,
Given the following binary tree,
         1
       /  \
      2    3
     / \    \
    4   5    7
After calling your function, the tree should look like:
         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

@tag-tree

Algorithm

DFS

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.

BFS

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

Code

Java

  • 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
                    }
                }
                return real_root;
            }
        }
    
    
    }
    
    
  • // 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 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
    
    

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