117. Populating Next Right Pointers in Each Node II

题目描述

题目链接

https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

方法思路

Approach1:

class Solution {
    //Runtime: 0 ms, faster than 100.00%
    //Memory Usage: 52.9 MB, less than 38.95% 
    public Node connect(Node root) {
        Node dummyHead = new Node(0);
        Node pre = dummyHead, cur = root;
        while (cur != null) {
	        if (cur.left != null) {
		        pre.next = cur.left;
		        pre = pre.next;
	        }
	        if (cur.right != null) {
		        pre.next = cur.right;
		        pre = pre.next;
	        }
	        cur = cur.next;
	        if (cur == null) {
		        pre = dummyHead;
		        cur = dummyHead.next;
		        dummyHead.next = null;
                //This line of code is for the bottom level. 
                //If you get rid of this line, you will get a TLE because 
                //there is a  dead loop in the bottom level.
	        }
        }
        return root;
    }
}

Approach2:利用队列的数据结构+基于层序遍历

class Solution {
    //Runtime: 2 ms, faster than 70.08%
    //Memory Usage: 51.6 MB, less than 42.59% 
    public Node connect(Node root) {
        if(root == null) return root;
        Queue<Node> q1 = new LinkedList<>();
        q1.offer(root);
        while(!q1.isEmpty()){
            int size = q1.size();
            Node node = q1.poll(), cur = node;
            while(size-- > 1){
                if(node.left != null) q1.offer(node.left);
                if(node.right != null) q1.offer(node.right);
                cur = q1.poll();
                node.next = cur;
                node = cur;
            }
            if(node.left != null) q1.offer(node.left);
            if(node.right != null) q1.offer(node.right);
        }            
        return root;
    }
}

Approach3:基于层序遍历

class Solution {
    //Runtime: 1 ms, faster than 88.06% 
    //Memory Usage: 51.5 MB, less than 43.05% 
    //based on level order traversal
    public Node connect(Node root) {
        Node head = null; //head of the next level
        Node prev = null; //the leading node on the next level
        Node cur = root;  //current node of current level
        while (cur != null) {
            while (cur != null) { //iterate on the current level
                //left child
                if (cur.left != null) {
                    if (prev != null) {
                        prev.next = cur.left;
                    } else {
                        head = cur.left;
                    }
                    prev = cur.left;
                }
                //right child
                if (cur.right != null) {
                    if (prev != null) {
                        prev.next = cur.right;
                    } else {
                        head = cur.right;
                    }
                    prev = cur.right;
                }
                //move to next node
                cur = cur.next;
            }
            //move to next level
            cur = head;
            head = null;
            prev = null;
        }
        return root;
    }
}