[leedcode 117] 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

/**
 * 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) {
        //注意最后访问的顺序，先右侧再左侧，因为在找p时，需要右侧p能够找到最左节点，例子{2,1,3,0,7,9,1,2,#,1,0,#,#,8,8,#,#,#,#,7}，
        //7->9->1->8
        //本题和之前的不同是，找节点的next时，需要根节点的next，甚至next的next
        if(root==null) return;
        TreeLinkNode p=root.next;
        while(p!=null){
            if(p.left!=null){
                p=p.left;
                break;
            }
            if(p.right!=null){
                p=p.right;
                break;
            }
            p=p.next;
            
        }
        if(root.left!=null){
            root.left.next=root.right==null?p:root.right;
        }
        if(root.right!=null){
            root.right.next=p;
        }
        connect(root.right);
        connect(root.left);//注意顺序
       
        
    }
}