Populating Next Right Pointers in Each Node II 任意（非完美）二叉树添加next指针 @LeetCode

http://blog.csdn.net/fightforyourdream/article/details/14514165

在上题基础上扩展，此时数不是完美二叉树，而是任意二叉树

这道题有些tricky，有时间回头再研究一下

package Level4;

import Utility.TreeLinkNode;

/**
* Populating Next Right Pointers in Each Node II
*
*  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:

You may only use constant extra space.
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
Discuss

*
*/
public class S117 {

public static void main(String[] args) {

}

public static void connect(TreeLinkNode root) {
// 空节点就直接返回
if (root == null){
return;
}

// 找到与root同一行的next node
TreeLinkNode rootNext = root.next;
TreeLinkNode next = null;		// 下一个被连接的对象

// rootNext如果是null说明已经处理完这一层的所有node
// next不等于null说明找到了找到最左边的下一个被连接的对象
while (rootNext != null && next == null)
{
if (rootNext.left != null){	// 优先找左边
next = rootNext.left;
} else{
next = rootNext.right;
}
rootNext = rootNext.next;
}

if (root.left != null)
{
if (root.right != null){	//	内部相连
root.left.next = root.right;
}else{						// 跨树相连
root.left.next = next;
}
}
if (root.right != null){		// 跨树相连
root.right.next = next;
}

connect(root.right);		// 要先让右边都先连起来
connect(root.left);
}
}

第二次做修改了一个地方的bug（在找rootnext时增加了break，及时退出），居然提交两次就通过了。

果然做LeetCode的真谛在于：遇到不懂 => 查答案 => 自己理解写出来或背下来 => 过一段时间后第二遍重做

比如这道题，我印象最深刻的是一定要先处理右子树再处理左子树，知道这点就足够写出来了。

关键就是找到next节点是在哪里，而这又要求找到rootnext节点在哪里。那个while循环就是破题的关键。

/**
* Definition for binary tree with next pointer.
* public class TreeLinkNode {
* int val;
* TreeLinkNode left, right, next;
* TreeLinkNode(int x) { val = x; }
* }
*/
public class Solution {
public void connect(TreeLinkNode root) {
if(root == null){
return;
}

TreeLinkNode rootnext = root.next;
TreeLinkNode next = null;
while(rootnext != null){
if(rootnext.left != null){
next = rootnext.left;
break;
}else if(rootnext.right != null){
next = rootnext.right;
break;
}else{
rootnext = rootnext.next;
}
}

if(root.right != null){
root.right.next = next;
}
if(root.left != null){
if(root.right != null){
root.left.next = root.right;
}else{
root.left.next = next;
}
}

connect(root.right);
connect(root.left);
}
}