leetcode No102. Binary Tree Level Order Traversal

Datetime:2016-08-23 01:17:06          Topic: LeetCode           Share

Question:

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:

Given binary tree  [3,9,20,null,null,15,7] ,

3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

一层一层的输出

Algorithm:

用一个count记录层数,具体见程序

Accepted Code:

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        if(root==NULL)return res;
        queue<TreeNode*> q;
        vector<int> temp;
        q.push(root);
        int curCount=1;
        int nextCount=0;
        while(!q.empty())
        {
            TreeNode* curNode=q.front();
            temp.push_back(curNode->val);
            q.pop();
            curCount--;
            if(curNode->left)
            {
                q.push(curNode->left);
                nextCount++;
            }
            if(curNode->right)
            {
                q.push(curNode->right);
                nextCount++;
            }
            if(curCount==0)
            {
                curCount=nextCount;
                nextCount=0;
                res.push_back(temp);
                temp.clear();
            }
        }
        return res;
    }
};




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