Follow up for problem "Populating Next Right Pointers in Each Node".
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
For example,
Given the following binary tree,
1
/ 2 3
/ \ 4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ 2 -> 3 -> NULL
/ \ 4-> 5 -> 7 -> NULL
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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){ //looking for the head of next level while(topLevel != null
&& nextLevelHead == null){ if(topLevel.left != null){ nextLevelHead = topLevel.left; node = nextLevelHead; } if(topLevel.right != null){ if(node == null){ nextLevelHead = topLevel.right; node = nextLevelHead; } else{ node.next = topLevel.right; node = node.next; } } topLevel = topLevel.next; } while(topLevel != null){ if(topLevel.left != null){ node.next = topLevel.left; node = node.next; } if(topLevel.right != null){ 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 II,布布扣,bubuko.com
leetcode--Populating Next Right Pointers in Each Node II
原文地址：http://www.cnblogs.com/averillzheng/p/3769567.html
