LeetCode

102. Binary Tree Level Order Traversal

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

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

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

层次遍历，用两个队列实现的。时间复杂度O(n)，空间复杂度O(n)

评论区有用一个队列实现的，是在每层后面加一个NULL，只要扫到NULL就说明换层，嗯，靠谱

/** * 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*> qu;        queue<TreeNode*> last;        qu.push(root);        TreeNode* pre = root;        vector<int> miao;        while (true) {            bool flag = false;;            while (!qu.empty()) {                flag = true;                TreeNode* cur = qu.front();                miao.push_back(cur->val);                last.push(cur);                qu.pop();            }            if (flag) {                ans.push_back(miao);                miao.clear();            }            while (!last.empty()) {                flag = true;                TreeNode* cur = last.front();                if (cur->left) qu.push(cur->left);                if (cur->right) qu.push(cur->right);                last.pop();            }            if (!flag) break;        }        return ans;    }};

class Solution {public:    vector<vector<int> > levelOrder(TreeNode *root) {        vector<vector<int> >  result;        if (!root) return result;        queue<TreeNode*> q;        q.push(root);        q.push(NULL);        vector<int> cur_vec;        while(!q.empty()) {            TreeNode* t = q.front();            q.pop();            if (t==NULL) {                result.push_back(cur_vec);                cur_vec.resize(0);                if (q.size() > 0) {                    q.push(NULL);                }            } else {                cur_vec.push_back(t->val);                if (t->left) q.push(t->left);                if (t->right) q.push(t->right);            }        }        return result;    }};

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]]

层次遍历倒着输出，跟上边那题一样。这次借鉴了上一题评论区的做法，采用一个NULL指针放入每一层的末尾，只需一个队列即可。时间复杂度O(n)，空间复杂度O(n)

/** * 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>> levelOrderBottom(TreeNode* root) {        vector<vector<int> >ans;        if (!root) return ans;        queue<TreeNode*> q;        q.push(root);        q.push(NULL);        vector<int> miao;        while (!q.empty()) {            TreeNode* cur = q.front();            q.pop();            if (cur == NULL) {                ans.insert(ans.begin(), miao);                miao.clear();                if (!q.empty()) q.push(NULL);            }            else {                miao.push_back(cur->val);                if (cur->left) q.push(cur->left);                if (cur->right) q.push(cur->right);            }        }        return ans;    }};

637. Average of Levels in Binary Tree

Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

Example 1:

Input:    3   / \  9  20    /  \   15   7Output: [3, 14.5, 11]Explanation:The average value of nodes on level 0 is 3,  on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].

Note:

1. The range of node's value is in the range of 32-bit signed integer.

额，跟上边两道没啥不一样的，就是求个平均数。时间复杂度O(n)，空间复杂度O(n)

/** * 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<double> averageOfLevels(TreeNode* root) {        vector<double> ans;        if (!root) return ans;        queue<TreeNode*> q;        q.push(root);        q.push(NULL);        double sum = 0.0;        int cnt = 0;        while (!q.empty()) {            TreeNode* cur = q.front();            q.pop();            if (cur == NULL) {                ans.push_back(sum * 1.0 / cnt);                sum = 0;                cnt = 0;                if (!q.empty()) q.push(NULL);            }            else {                sum += cur->val;                cnt++;                if (cur->left) q.push(cur->left);                if (cur->right) q.push(cur->right);            }        }        return ans;    }};