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leetcode--Populating Next Right Pointers in Each Node

2024-07-07 03:42:47   220人阅读

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

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/**
 * 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

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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;
            }
        }
    }
}

　　　　


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