102. 二叉树的层序遍历

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int fun(TreeNode* root)
    {
        if(root==nullptr)
        {
            return 0;
        }
        int left=fun(root->left);
        int right=fun(root->right);
        return 1+(left>right?left:right);
    }
    vector<vector<int>> levelOrder(TreeNode* root) {
        if(root==nullptr)
            return {};
        int h =fun(root);
        vector<vector<int>>v(h, vector<int>());
        queue<TreeNode*>q;
        q.push(root);
        int i=0;

        //记录上一层节点的数量
        int pre=1;
        int cnt=0;
        // 遇到的难点1、怎么确定每层的节点
        while(!q.empty())
        {
            TreeNode* node = q.front();
            if(pre==0)
            {
                pre=cnt;
                i++;
                cnt=0;
            }
            v[i].push_back(node->val);
            q.pop();
            pre--;
            if(node->left)
            {
                q.push(node->left);
                cnt++;
            }
                
            if(node->right)
            {
                q.push(node->right);
                cnt++;
            }
        }

        return v;
    }
};

简洁写法

双队列解法

一个队列存节点指针
一个队列存层数

还是要求深度，要不还是不大好搞

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class Solution {
public:

    int fun(TreeNode* root)
    {
        if(root==nullptr)
        {
            return 0;
        }
        int left=fun(root->left);
        int right=fun(root->right);
        return 1+(left>right?left:right);
    }
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<int>level;
        queue<TreeNode*>q;

        if(root)
        {
            q.push(root);
            level.push(0);
        }

        int h=fun(root);
        vector<vector<int>>vv(h,vector<int>());
        while(!q.empty())
        {
            TreeNode* front = q.front();
            int pre=level.front();
            q.pop();
            level.pop();

            vv[pre].push_back(front->val);

            if(front->left)
            {
                q.push(front->left);
                level.push(pre+1);
            }

            if(front->right)
            {
                q.push(front->right);
                level.push(pre+1);
            }
        }

        return vv;
    }
};