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Populating Next Right Pointers in Each Node II

时间:2014-08-20 22:47:33      阅读:297      评论:0      收藏:0      [点我收藏+]

标签:leetcode   遍历   traversal   

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

使用一个队列保存当前节点和当前层号,prev保存这个节点的前一个节点,如果节点属于同一层,则prev的next指向队列的头元素。

层次遍历

Have you been asked this question in an interview? 
/**
 * 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(!root) return;
        queue< pair<TreeLinkNode*,int> > q;
        q.push(pair<TreeLinkNode*,int>(root,0));
        pair<TreeLinkNode* ,int> prev;
        while(!q.empty()){
            pair<TreeLinkNode*,int> temp=q.front();
            q.pop();
            if(prev.first!=NULL&& prev.second==temp.second)
                prev.first->next=temp.first;
            prev=make_pair(temp.first,temp.second);
            if(temp.first->left) q.push(pair<TreeLinkNode*,int>(temp.first->left,temp.second+1));
            if(temp.first->right) q.push(pair<TreeLinkNode*,int>(temp.first->right,temp.second+1));
        }
    }
};


Populating Next Right Pointers in Each Node II,布布扣,bubuko.com

Populating Next Right Pointers in Each Node II

标签:leetcode   遍历   traversal   

原文地址:http://blog.csdn.net/huangynn/article/details/38712315

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