[LeetCode] Populating Next Right Pointers in Each Node I, II

題目：

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,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

 

After calling your function, the tree should look like:

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

分析：

這道題之所以放上來是因為題目中的那句話：You may only use constant extra space

這就意味着，深搜是不能用的，因為遞歸是需要棧的，因此空間復雜度將是 O(logn)。毫無疑問廣搜也不能用，因為隊列也是占用空間的，空間占用還高於 O(logn)

 

難就難在這里，深搜和廣搜都不能用，怎么完成樹的遍歷？

我拿到題目的第一反應便是：用廣搜，接着發現廣搜不能用，便犯了難。

看了一些提示，有招了：核心仍然是廣搜，但是我們可以借用 next 指針，做到不需要隊列就能完成廣度搜索。

 

如果當前層所有結點的next 指針已經設置好了，那么據此，下一層所有結點的next指針 也可以依次被設置。

代碼：

/**
 * 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,

         1
       /  \
      2    3
     / \    \
    4   5    7

 

After calling your function, the tree should look like:

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

隨后題目做了一些更改：不一定是滿二叉樹。

 

解法的核心：遞推思想 依然不需要改變，依然是依據當前層的next 指針，設置下一層的 next 指針。只是找結點麻煩些，我們定義了兩個函數，findNextNodeNextLev用來找(n+1)層的下一個節點，findStartNodeNextLev 用來找下一層的起始節點。

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);
    }
};