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

本文介绍了一种在任意二叉树中填充每个节点的Next指针的方法，确保相同层级的相邻节点通过Next指针相连。该算法仅使用常数额外空间，并详细解释了如何遍历树及更新Next指针。

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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,

After calling your function, the tree should look like:

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

是前一道题的延伸，这里树不再是complete的了，而可以是任意形式的。

需要做些改动。在找下一层的head pointer时要判断current有哪个孩子；而且也不一定是记录前一个node的right child了，而是记录最近的一个没有更新其next的node。

/**
 * 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 nextLineHead = root;
        
        while(nextLineHead != null){
            TreeLinkNode previousTail = null, current = nextLineHead;
            nextLineHead = null;
            
            while(current != null){
                //update the next line header if the current node has a child
                if(nextLineHead == null){
                    if(current.left != null)
                        nextLineHead = current.left;
                    else if(current.right != null)
                        nextLineHead = current.right;
                }
                
                //if the current node has two children, link the left to the right
                if(current.left != null && current.right != null)
                    current.left.next = current.right;
                
                //link the previousTail node to one of current node's children if possible
                if(previousTail != null){
                    if(current.left != null && current.right != null){
                        previousTail.next = current.left;
                        previousTail = current.right;
                    }
                    else if(current.left != null){
                        previousTail.next = current.left;
                        previousTail = current.left;
                    }
                    else if(current.right != null){
                        previousTail.next = current.right;
                        previousTail = current.right;
                    }
                }
                else if(nextLineHead != null){
                    if(nextLineHead.next != null)
                        previousTail = nextLineHead.next;
                    else
                        previousTail = nextLineHead;
                }
                //move to the next node
                current = current.next;
            }
        }
    }
    
}