Populating Next Right Pointers in Each Node I, II

最新推荐文章于 2020-11-14 09:32:10 发布

Algebra2016

最新推荐文章于 2020-11-14 09:32:10 发布

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本文链接：https://blog.youkuaiyun.com/Algebra2016/article/details/78382958

本文探讨了一种在常数额外空间限制下，利用next指针进行二叉树遍历的方法，特别适用于完美二叉树及一般二叉树场景。

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题目：

Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

分析：

这道题之所以放上来是因为题目中的那句话：You may only use constant extra space

这就意味着，深搜是不能用的，因为递归是需要栈的，因此空间复杂度将是 O(logn)。毫无疑问广搜也不能用，因为队列也是占用空间的，空间占用还高于 O(logn)

难就难在这里，深搜和广搜都不能用，怎么完成树的遍历？

我拿到题目的第一反应便是：用广搜，接着发现广搜不能用，便犯了难。

看了一些提示，有招了：核心仍然是广搜，但是我们可以借用 next 指针，做到不需要队列就能完成广度搜索。

如果当前层所有结点的next 指针已经设置好了，那么据此，下一层所有结点的next指针也可以依次被设置。

代码：

/**
 * 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) {
        if(NULL == root) return;
        TreeLinkNode* curLev;
        while(root -> left != NULL){
            curLev = root;
            while(curLev != NULL){
                curLev -> left -> next = curLev -> right;
                if(curLev -> next != NULL)
                    curLev -> right -> next = curLev -> next -> left;
                curLev = curLev -> next;
            }
            root = root -> left;
        }
    }
};

引申：

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

随后题目做了一些更改：不一定是满二叉树。

解法的核心：递推思想依然不需要改变，依然是依据当前层的next 指针，设置下一层的 next 指针。只是找结点麻烦些，我们定义了两个函数，findNextNodeNextLev用来找(n+1)层的下一个节点，findStartNodeNextLev 用来找下一层的起始节点。

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(NULL == root) return;
        TreeLinkNode* start;
        TreeLinkNode* curNode;
        TreeLinkNode* nextNode;
        while(root != NULL){
            start = findStartNodeNextLev(root);
            curNode = start;
            nextNode = findNextNodeNextLev(root, start);
            while(nextNode != NULL){
                curNode -> next = nextNode;
                curNode = nextNode;
                nextNode = findNextNodeNextLev(root, curNode);
            }
            root = start;
        }
    }
private:
    TreeLinkNode* findNextNodeNextLev(TreeLinkNode* &cur, TreeLinkNode* curNextLev){
        if(cur -> left == curNextLev && cur -> right != NULL){
            return cur -> right;
        }else{
            while(cur -> next != NULL){
                cur = cur -> next;
                if(cur -> left != NULL && cur -> left != curNextLev) return cur -> left;
                if(cur -> right != NULL && cur -> right != curNextLev) return cur -> right;
            }
        }
        return NULL;
    }
    
    TreeLinkNode* findStartNodeNextLev(TreeLinkNode* node){
        if(NULL == node) return NULL;
        if(node -> left != NULL) return node -> left;
        return findNextNodeNextLev(node, node -> left);
    }
};