Populating Next Right Pointers in Each Node II -- leetcode
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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
基本思路:
和 Populating Next Right Pointers in Each Node 一样,以当前结点为一个处理单位,
1. 将右子树的结点挂到左子树的next上;
2. 将右邻居的左子树结点,持到当前结点right子树上的next上。
但,
和 Populating Next Right Pointers in Each Node 相比,缺少的条件是,不再是一个完全二叉树了。
不能简单的通过从root结点出发,延着left指针,找到每层链表头了。
故当完成上一个层次的链接后,还需要一定的措施,找到下一层的起始点。
方法之一,就是延着上一层的next指针,找到第一个left 或者 right不为空的结点。顺着该结点,就找到了下一层起始点。
另外一个不同点是,将右邻居的left子树,挂到当前结点的right子树的next上时,右邻居,可能没有left,甚至也没有right。那么就只能延着next一种搜索下去,找到第一个有孩子的结点。将其孩子串到当前结点的子树的next上。
此代码在leetcode上实际执行时间为43ms。
/** * 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) { while (root && (root->left || root->right)) { TreeLinkNode *p = root; while (p) { TreeLinkNode *tail = 0; if (p->left && p->right) { p->left->next = p->right; tail = p->right; } else if (p->left) tail = p->left; else tail = p->right; TreeLinkNode *q = p->next; while (q && !q->left && !q->right) q = q->next; if (q) tail->next = q->left ? q->left : q->right; p = q; } root = root->left ? root->left: root->right; while (root && !root->left && !root->right) root = root->next; } }};
上面的写法还是点麻烦,
在内层循环中,建立一个伪链表头,指向树下一层串起来链表的表头。
当完成该层串接后,只要通过该链表头,就可以找到向一层的起始结点。
在内层循环过层中,维持一指向链表尾元素的指针,可以方便的把left 和右子树串起来。变成了一个普通的链表尾部的添加元素操作。
在代码在leetcode上实际执行时间为38ms。class Solution {public: void connect(TreeLinkNode *root) { while (root) { TreeLinkNode head(0); TreeLinkNode *tail = &head; while (root) { if (root->left) { tail->next = root->left; tail = tail->next; } if (root->right) { tail->next = root->right; tail = tail->next; } root = root->next; } root = head.next; } }
此代码参考自:
https://leetcode.com/discuss/24398/simple-solution-using-constant-space
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