[leetcode]: 107. Binary Tree Level Order Traversal II

1.题目

Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

 3/ \9  20/  \15  7

return its bottom-up level order traversal as:
[[15,7],[9,20],[3]]
二叉树按层来遍历，自底向上

2.分析

BFS：广度优先遍历，用vector存每层遍历的结果。最后reverse
DFS：深度优先遍历，需要加额外参数level来标记当前元素在哪一层。

3.代码

BFS

class Solution {public:    vector<vector<int>> levelOrderBottom(TreeNode* root) {        vector<vector<int>> ans;        if (root == NULL)            return ans;        //广度优先用queue，深度优先用stack        queue<TreeNode*> nodes;        nodes.push(root);        while (!nodes.empty()) {            int num = nodes.size();            vector<int> level;            for (int i = 0; i < num; i++) {                TreeNode* tmp = nodes.front();                nodes.pop();                level.push_back(tmp->val);                if (tmp->left != NULL)                    nodes.push(tmp->left);                if (tmp->right != NULL)                    nodes.push(tmp->right);                     }            ans.push_back(level);        }        reverse(ans.begin(),ans.end());        return ans;    }};

DFS

class Solution {public:    void DFS(vector<vector<int>>& ans,TreeNode* root, int level) {        if (root == NULL)            return;        if (level >= ans.size())            ans.push_back(vector<int>());        ans[level].push_back(root->val);        DFS(ans, root->left, level + 1);        DFS(ans, root->right, level + 1);    }    vector<vector<int>> levelOrderBottom(TreeNode* root) {        vector<vector<int>> ans;        if (root == NULL)            return ans;        DFS(ans, root, 0);        reverse(ans.begin(), ans.end());        return ans;    }};