[leetcode] Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-09-24 12:05:40 发布

Cabinathor

最新推荐文章于 2021-09-24 12:05:40 发布

阅读量346

点赞数

CC 4.0 BY-SA版权

分类专栏： Leetcode

本文链接：https://blog.youkuaiyun.com/lmxmimihuhu/article/details/46642801

Leetcode 专栏收录该内容

222 篇文章

订阅专栏

本文讨论了如何在不使用额外空间的情况下，解决任意二叉树中节点的链接问题，通过两种优化方法实现从上层到下层节点的连接。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

From : https://leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/

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

Solution 1: 每次从上一层取到下一层的下一个。

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        bool isLeft=false;
        TreeLinkNode *cur=root, *up, *dp, *t=NULL;
        while(cur) {
            up = cur;
            while(up) {
                if(!up->left && !up->right) {
                    up=up->next;
                    continue;
                }
                if(up->left) {
                    isLeft = true;
                } else {
                    isLeft = false;
                } 
                if(!t) {
                    dp = t = isLeft ? up->left : up->right;
                    if(isLeft && up->right) {
                        dp->next = up->right;
                        dp = dp->next;
                    }
                    up=up->next;
                    continue;
                }
                dp->next = isLeft ? up->left : up->right;
                dp = dp->next;
                if(isLeft && up->right) {
                    dp->next = up->right;
                    dp = dp->next;
                }
                up=up->next;
            }
            cur = t;
            t = NULL;
        }
    }
};

Solution 2:

简化为：

/**
 * Definition for binary tree with next pointer.
 * struct TreeLinkNode {
 *  int val;
 *  TreeLinkNode *left, *right, *next;
 *  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
 * };
 */
class Solution {
public:
    void connect(TreeLinkNode *root) {
        TreeLinkNode* prev = NULL, *nextLevel = NULL;
        
        while(root) {
            prev = nextLevel = NULL;
            while(root) {
                if(nextLevel == NULL) nextLevel = root->left ? root->left : root->right;
                if(root->left) {
                    if(prev) prev->next = root->left;
                    prev = root->left;
                }
                if(root->right) {
                    if(prev) prev->next = root->right;
                    prev = root->right;
                }
                root = root->next;
            }
            root = nextLevel;
        }
    }
};

/**
 * 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) {
		TreeLinkNode up = root, cur = null, newLevel = null;

		while (up != null) {
			cur = newLevel = null;
			while (up != null) {
				if (newLevel == null) {
					// 没初始化的时候 想办法初始化
					newLevel = up.left != null ? up.left : up.right;
				}
				if (up.left != null) {
					if (cur != null) {
						cur.next = up.left;
					}
					cur = up.left;
				}
				if (up.right != null) {
					if (cur != null) {
						cur.next = up.right;
					}
					cur = up.right;
				}
				up = up.next;
			}
			up = newLevel;
		}
    }
}