【LeetCode】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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 return its level order traversal as:

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

提示：

此题要求逐行输出二叉树的节点数值，这里提供两种解法，第一种基于BFS，第二种基于DFS。

代码：

BFS：

/**
 * 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) return res;
        queue<TreeNode*> q;
        vector<int> v;
        q.push(root);
        q.push(NULL);
        TreeNode *node;
        while(!q.empty()) {
            node = q.front();
            q.pop();
            if (node == NULL) {
                res.push_back(v);
                v.clear();
                if (q.size()) q.push(NULL);
            } else {
                v.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
        }
        return res;
    }
};

 DFS:

/**
 * 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) {
        search(root, 0);
        return res;
    }
private:
    void search(TreeNode* node, int depth) {
        if (!node) return;
        if (res.size() == depth) {
            res.push_back(vector<int>());
        }
        
        res[depth].push_back(node->val);
        search(node->left, depth + 1);
        search(node->right, depth + 1);
    }
    vector<vector<int>> res;
};