關於二叉樹的遍歷查詢。

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]
]

/**
 * 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 = new vector<vector<int>>();
        
        levelHelper(res,root,0);      
        
        return *res;        
    }
    //採用DFS的方法求解
    void levelHelper(vector<vector<int>>* res,TreeNode* root,int level)
    {
        if(root == NULL)  return;
        
        if(level >= res->size())
        {
            res->push_back(vector<int>());
        }
        // cout<<"root_val="<<root->val<<"  level="<<level<<endl;
        res->at(level).push_back(root->val);
        
        levelHelper(res,root->left,level+1);
        levelHelper(res,root->right,level+1);
        
    }
};
 

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解法參考連結