Binary Tree Level Order Traversal

给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。

例如:
给定二叉树: [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

返回其层次遍历结果：

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

1、递归版

/**
 * 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;
        levelorder(root, 0, result);
        return result;
    }
    void levelorder(TreeNode* root, int level, vector<vector<int>>& result)
    {
        
        if (!root)
            return;
        
        if(level == result.size())
            result.push_back({});
        result[level].push_back(root->val);
        if(root->left)levelorder(root->left, level+1, result);
        if(root->right)levelorder(root->right, level+1, result);
    }
};

 

2、迭代版

/**
 * 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 == NULL) return res;

        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            vector<int> oneLevel;
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                TreeNode *node = q.front();
                q.pop();
                oneLevel.push_back(node->val);
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            res.push_back(oneLevel);
        }
        return res;
    }
    
};

 

 参考：https://www.cnblogs.com/grandyang/p/4051321.html