LeetCode - Populating Next Right Pointers in Each Node II-优快云博客

本文探讨了如何通过递归方法解决任意二叉树中节点的next指针问题，并实现了常数额外空间的解决方案。具体步骤包括连接root的左右节点，并优先处理右树，最后连接左树。

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题目：

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,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

思路：

递归，对root的左右进行连接操作，然后先递归右树，再递归左树。

package tree;

public class PopulatingNextRightPointersInEachNodeII {

    public void connect(TreeLinkNode root) {
        if (root == null || (root.left == null && root.right == null)) return;

        TreeLinkNode next = root.next;
        while (next != null) {
            if(next.left != null) {
                next = next.left;
                break;
            }
            
            if (next.right != null) {
                next = next.right;
                break;
            }
            
            next = next.next;
        }
        
        if (root.right != null) 
            root.right.next = next;
         
        if (root.left != null)
            root.left.next = root.right == null ? next : root.right;
                
        connect(root.right);
        connect(root.left);
    }

}