利用bds和dfs解决 LeetCode 107. Binary Tree Level Order Traversal II

问题简述

给定一棵二叉树，返回该二叉树自底向上遍历的结点值（即从左到右，自底向上）
比如给定一颗二叉树 [3,9,20,null,null,15,7]

    3   / \  9  20    /  \   15   7

返回的结果为

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

解决方案

解法1：广度优先遍历（BFS）

最简单的想法就是先广度优先遍历，按层存值，最后倒一下~
直接上代码

/** * 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 {private:    vector<vector<int>> res;private:    void _bfs(TreeNode* root){        queue<TreeNode*> q;        q.push(root);        while (!q.empty()){            int len = q.size();            vector<int> vec;            for (int i = 0; i < len; i++){                TreeNode* tmp = q.front();                q.pop();                if (tmp->left)                    q.push(tmp->left);                if (tmp->right)                    q.push(tmp->right);                vec.push_back(tmp->val);            }            res.push_back(vec);        }    }public:    vector<vector<int>> levelOrderBottom(TreeNode* root) {        res.clear();        if (!root)            return res;        _bfs(root);        reverse(res.begin(), res.end());        return res;    }};

解法2：深度优先遍历（DFS）

另一种做法是，一边深度遍历，一边按照层数记录数据

/** * 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 {private:    vector<vector<int>> res;private:    void _dfs(TreeNode* root, int layer){        if (!root)            return;        if (layer >= res.size()){            res.insert(res.begin(), vector<int>());        }        res[res.size() - layer - 1].push_back(root->val);        _dfs(root->left, layer + 1);        _dfs(root->right, layer + 1);    }public:    vector<vector<int>> levelOrderBottom(TreeNode* root) {        res.clear();        if (!root)            return res;        _dfs(root, 0);        return res;    }};