Populating Next Right Pointers in Each Node II

本文探讨了如何在任意二叉树结构中填充节点的next指针,通过遍历顺序和while循环实现解决方案,同时强调了使用常数额外空间的重要性。示例展示了将树结构转换为具有next链接的链式结构的过程。

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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);
    }
}


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