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Populating Next Right Pointers in Each Node II

时间:2014-12-03 21:10:56      阅读:139      评论:0      收藏:0      [点我收藏+]

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
/**
 * 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 *head = NULL; //head of the next level
        TreeLinkNode *prev = NULL; //the leading node on the next level
        TreeLinkNode *cur = root;  //current node of current level

        while (cur != NULL) {

            while (cur != NULL) { //iterate on the current level
                //left child
                if (cur->left != NULL) {
                    if (prev != NULL) {
                        prev->next = cur->left;
                    } else {
                        head = cur->left;
                    }
                    prev = cur->left;
                }
                //right child
                if (cur->right != NULL) {
                    if (prev != NULL) {
                        prev->next = cur->right;
                    } else {
                        head = cur->right;
                    }
                    prev = cur->right;
                }
                //move to next node
                cur = cur->next;
            }

            //move to next level
            cur = head;
            head = NULL;
            prev = NULL;
        }
    }
};

 

Populating Next Right Pointers in Each Node II

原文:http://www.cnblogs.com/code-swan/p/4141155.html

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