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
Analysis:
1. Traversal order, root, right, left.
2. In order to populate the next right pointers, a while loop is a must. Especially, the order for checking the pointers is (root.right, nextnode.left), (root.right, nextnode.right), (root.left, nextnode.left), (root.right, nextnode.right).
/**
* 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;
if(root.left!=null && root.right!=null) root.left.next=root.right;
TreeLinkNode node = root.next;
while(node!=null) {
if(root.right!=null) {
if(node.left!=null) {
root.right.next=node.left;
break;
}
else if(node.right!=null) {
root.right.next=node.right;
break;
}
}
else if(root.left!=null) {
if(node.left!=null) {
root.left.next=node.left;
break;
}
else if(node.right!=null) {
root.left.next=node.right;
break;
}
}
node = node.next;
}
connect(root.right);
connect(root.left);
}
}