leetcode--Populating Next Right Pointers in Each Node

时间:2014-06-05 13:42:24   收藏:0   阅读:264

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:

 

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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