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; } } }} |
leetcode--Populating Next Right Pointers in Each Node,布布扣,bubuko.com
leetcode--Populating Next Right Pointers in Each Node
原文：http://www.cnblogs.com/averillzheng/p/3769124.html