LeetCode 107. Binary Tree Level Order Traversal II 树的BFS、DFS

107. Binary Tree Level Order Traversal II

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]
]

题意

类比102. Binary Tree Level Order Traversal

思路

102. Binary Tree Level Order Traversal 的结果逆序输出，这里使用方向迭代器
无论是102还是107都可以使用BFS和DFS两种方向解决。
达到遍历整颗树的效果，重点是维护好各个结点所属层数的信息。

代码

BFS

//bfsclass Solution {public:    vector<vector<int>> levelOrder(TreeNode* root) {        vector<vector<int>> res;         if(root == NULL)        {            return res;        }        //queue<int> que;  //想清楚队列存储的数据，是树的结点入队        //观察返回信息，还需包含层级信息，所以结点入队还需维护其层级的信息        queue<pair<TreeNode*,int>> que;        que.push(make_pair(root,0));        while(!que.empty())        {            //取队头元素            TreeNode *node = que.front().first;            int level = que.front().second;            que.pop();            //需要注意当前节点所在的层，是否在res中已经存在            if(level == res.size())  //若相等就说明res还不包含level层，这个结点肯定在新的层中                res.push_back(vector<int>());  //创建一个新的层,来加载该层的结点数            res[level].push_back(node->val);            if(node->left)            {                  que.push(make_pair(node->left,level+1));                //当左结点存在，那么左结点的层数是当前结点层数+1            }            if(node->right)            {                que.push(make_pair(node->right,level+1));            }        }        return vector<vector<int>>(res.rbegin(),res.rend());    }};

DFS

class Solution {public:    vector<vector<int> > res;    void DFS(TreeNode* root, int level)    {        if (root == NULL) return;        if (level == res.size()) // The level does not exist in output        {            res.push_back(vector<int>()); // Create a new level        }        res[level].push_back(root->val); // Add the current value to its level        DFS(root->left, level+1); // Go to the next level        DFS(root->right,level+1);    }    vector<vector<int> > levelOrderBottom(TreeNode *root) {        DFS(root, 0);        return vector<vector<int> > (res.rbegin(), res.rend());    }};