LeetCode OJ 102. Binary Tree Level Order Traversal

题目

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]
]

解答

其实就是个BFS。。。

下面是AC的代码：

/**
 * 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>> ans;
        vector<vector<TreeNode*>> temp;
        int cur_level = 0;
        if(root == 0){
            return ans;
        }
        ans.push_back(vector<int>());
        ans[0].push_back(root->val);
        temp.push_back(vector<TreeNode *>());
        temp[0].push_back(root);
        int next_level;
        int flag = 1;
        while(flag){
            flag = 0;
            next_level = cur_level + 1;
            ans.push_back(vector<int>());
            temp.push_back(vector<TreeNode *>());
            for(vector<TreeNode*>::iterator iter = temp[cur_level].begin(); iter != temp[cur_level].end(); iter++){
                if((*iter)->left != 0){
                    flag = 1;
                    ans[next_level].push_back((*iter)->left->val);
                    temp[next_level].push_back((*iter)->left);
                }
                if((*iter)->right != 0){
                    flag = 1;
                    ans[next_level].push_back((*iter)->right->val);
                    temp[next_level].push_back((*iter)->right);
                }
            }
            cur_level++;
        }
        ans.pop_back();
        return ans;
    }
};

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