LeetCode--Binary Tree Level Order Traversal

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]
]

方法一：遞歸。前向遍歷，先遍歷左節點，再遍歷右節點，每一層用一個動態數組存儲數據。

/**
 * 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>>result;
        traverse(root,1,result);
        return result;
    }
    void traverse(TreeNode* root,int level,vector<vector<int>>&result){
        if(!root) return;
        if(level>result.size()){
            result.push_back(vector<int>());
        }
        result[level-1].push_back(root->val);
        traverse(root->left,level+1,result);
        traverse(root->right,level+1,result);
    }
};

方法二：非遞歸。隊列，BFS.

class Solution {
public:
    vector<vector<int> > Print(TreeNode* pRoot) {
        vector<vector<int> > ret;
        queue<TreeNode*> que;
        if(pRoot) que.push(pRoot);
        while(!que.empty()){
            vector<int> tmp;
            int n = que.size();
            while(n--){
                TreeNode* node = que.front();
                tmp.push_back(node->val);
                if(node->left) que.push(node->left);
                if(node->right) que.push(node->right);
                que.pop();
            }
            ret.push_back(tmp);
        }
        return ret;
    }
};