Leetcode-117: Populating Next Right Pointers in Each Node II

最新推荐文章于 2025-08-08 00:00:33 发布

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最新推荐文章于 2025-08-08 00:00:33 发布

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分类专栏： leetcode 文章标签： leetcode 二叉树

本文链接：https://blog.youkuaiyun.com/cc5149/article/details/77428612

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本文介绍了一种在二叉树中填充每个节点的Next指针的方法，使得同一层的所有节点通过Next指针连接起来。文章提供了一个解决方案，能够在常数额外空间条件下完成任务，特别适用于非完全二叉树的情况。

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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

这道题和前面一题Populating Next Right Pointers in Each Node一样，唯一的不同在于，给出的不是一颗完全二叉树，不过解题思路依然是一样的，使用层序遍历。不过在遍历的时候需要考虑到的几种情况：1）h.left != null。此时，h.left.next的指向有四种情况：h.right, k.left, k.right, null，需要分析前三个节点是否为null；2）h.right != null。该节点的指向有三种情况：h.next.left, k.right, null。k为与h在同一层且在h之后，子节点至少有一个不为null的节点。

/**
 * 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 head = root, cur = null;
        while (head != null) {
            cur = head;
            boolean nextLevel = false;
            while (cur != null) {
                TreeLinkNode childNext = null;
                if (!nextLevel && (cur.left != null || cur.right != null)) {
                    nextLevel = true;
                    head = cur.left == null ? cur.right : cur.left;
                } 
                
                TreeLinkNode nextParent = cur.next;
                while (nextParent != null && nextParent.next != null && (nextParent.left == null && nextParent.right == null)){
                    nextParent = nextParent.next;
                }
                
                if (nextParent != null) {
                    childNext = nextParent.left == null ? nextParent.right : nextParent.left;
                }
                
                if (cur.right != null) {
                    if (cur.left != null)
                        cur.left.next = cur.right;
                    
                    cur.right.next = childNext;
                }
                else if (cur.left != null)
                    cur.left.next = childNext;
                
                cur = nextParent;
            }
            if (!nextLevel)
                break;
        }
    }
}