• leetcode--Populating Next Right Pointers in Each Node


    Given a binary tree

        struct TreeLinkNode {
          TreeLinkNode *left;
          TreeLinkNode *right;
          TreeLinkNode *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.

    Note:

    • You may only use constant extra space.
    • You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

    For example,
    Given the following perfect binary tree,

             1
           /  
          2    3
         /   / 
        4  5  6  7
    

    After calling your function, the tree should look like:

             1 -> NULL
           /  
          2 -> 3 -> NULL
         /   / 
        4->5->6->7 -> NULL

    /**
     * Definition for binary tree with next pointer.
     * public class TreeLinkNode {
     *     int val;
     *     TreeLinkNode left, right, next;
     *     TreeLinkNode(int x) { val = x; }
     * }
     */
    public class Solution {
        /**To solve this problem, we use a queue to save the nodes in  the same level.<br>
    	 * 
    	 * @param root -- The root node of the input tree.
    	 * @author Averill Zheng
    	 * @version 2014-06-03
    	 * @since JDK 1.7
    	 */
    	public void connect(TreeLinkNode root) {
    		if(root != null){
    			Queue<TreeLinkNode> node = new LinkedList<TreeLinkNode>();
    			node.add(root);
    			Queue<TreeLinkNode> nextLevel = new LinkedList<TreeLinkNode>();
    			while(node.peek() != null){
    				
    				TreeLinkNode first = node.poll();
    				if(first.left != null)
    					nextLevel.add(first.left);
    				if(first.right != null)
    					nextLevel.add(first.right);
    				
    				while(node.peek() != null){
    					TreeLinkNode aNode = node.poll();
    					if(aNode.left != null)
    						nextLevel.add(aNode.left);
    					if(aNode.right != null)
    						nextLevel.add(aNode.right);
    					first.next = aNode;
    					first = first.next;
    				}
    				first.next = null;
    				node = nextLevel;
    				nextLevel = new LinkedList<TreeLinkNode>();
    			}
    		}
        }
    }
    

    The following code uses constant extra memory

    public class Solution {
    	public void connect(TreeLinkNode root) {
    	    if(root != null){
    			root.next = null;
    			TreeLinkNode topLevel = root;
    			TreeLinkNode nextLevelHead = null;
    			TreeLinkNode node = null;
    			while(topLevel != null){
    				if(topLevel.left != null){
    					nextLevelHead = topLevel.left;
    					node = nextLevelHead;
    					node.next = topLevel.right;
    					node = node.next;
    				}
    				topLevel = topLevel.next;
    				while(topLevel != null){
    					if(topLevel.left != null && node != null){
    						node.next = topLevel.left;
    						node = node.next;
    						node.next = topLevel.right;
    						node = node.next;
    					}
    					topLevel = topLevel.next;
    				}
    				if(node != null){
    					node.next = null;
    					node = node.next;
    				}
    				topLevel = nextLevelHead;
    				nextLevelHead = null;
    			}
    		} 
        }
    }
    

        

  • 相关阅读:
    很实用的jQuery事件
    移动端touchstart,touchmove,touchend
    Django的学习(二)————Templates
    Django的学习(一)————初入django
    Tkinter添加图片
    HierSort(希尔)————Java
    Bubble(冒泡排序)————Java
    类+进程池的方法爬取喜马拉雅
    Ajax的爬取心得
    python中将两个数组压缩成一个数组
  • 原文地址:https://www.cnblogs.com/averillzheng/p/3769124.html
Copyright © 2020-2023  润新知