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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
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Initially, all next pointers are set to NULL
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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
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Code
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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; } }
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// 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; } };
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""" # 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
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/** * 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 }
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/** * 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; }