[Leetcode]102. 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]]

迭代：

/** * 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;        if (root == nullptr)            return result;                queue<TreeNode *> Q;        Q.push(root);        while (!Q.empty()) {            int size = Q.size();            vector<int> level;            for (int i = 0; i != size; ++i) {                TreeNode* head = Q.front();                Q.pop();                level.push_back(head->val);                if (head->left != nullptr)                    Q.push(head->left);                if (head->right != nullptr)                    Q.push(head->right);            }            result.push_back(level);        }        return result;    }};

递归：

/** * 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, size_t 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);    }};