LeetCode 102. 二叉树的层次遍历

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]
]

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/binary-tree-level-order-traversal

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:
    void process(TreeNode* root,int depth,vector<vector<int>> &ans)
    {

        if(depth >= ans.size())
            ans.push_back(vector<int> {});

        ans[depth].push_back(root->val);
        if(root->left) process(root->left,depth+1,ans);
        if(root->right) process(root->right,depth+1,ans);
    }

    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;

        if(root)
            process(root,0,ans);
        return ans;
    }   
};

 

非递归：

/**
 * 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>> ans;
        if(!root) return ans;

        queue<TreeNode*> Q;
        TreeNode* temp;
        Q.push(root);
        while (!Q.empty()){
            vector<int> ress;
            int n = Q.size();
            for(int i = 0; i < n; i++) {
                temp = Q.front();
                ress.push_back(temp->val);
                Q.pop();
                if(temp->left)
                    Q.push(temp->left);
                if(temp->right)
                    Q.push(temp->right);
            }
            ans.push_back(ress);
        }
        return ans;
    }
};