[Leetcode] 102, 107, 101

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]]

Solution: 先序遍历或者后序遍历都行，遍历的同时记录一下深度就可以了

Code（递归版）:

/** * 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;        travel(root, 0, ans);        return ans;    }private:    void travel(TreeNode* root, int depth, vector<vector<int>>& m){        if(root==NULL) return;        if(depth>=m.size()) m.push_back(vector<int>());        travel(root->left, depth+1, m);        travel(root->right, depth+1, m);        m[depth].push_back(root->val);    }};

Code（迭代版）: 使用queue记录每一层。

/** * 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;        queue<TreeNode*> cur,next;                if(root==NULL) return ans;        cur.push(root);        vector<int> levelval;        while(!cur.empty()){            while(!cur.empty()){                if(cur.front()->left!=NULL) next.push(cur.front()->left);                if(cur.front()->right!=NULL) next.push(cur.front()->right);                levelval.push_back(cur.front()->val);                cur.pop();            }            swap(cur,next);            ans.push_back(levelval);            levelval.clear();        }                return ans;    }};

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]]

Solution: 对前一题的结果进行翻转即可。

Code:

/** * 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;        getSons(root, 0, ans);        reverse(ans.begin(), ans.end()); //翻转        return ans;    }private:    void getSons(TreeNode* root, int d, vector<vector<int>>& v){        if(root==NULL) return;                if(d>=v.size()) v.push_back(vector<int>());        getSons(root->left, d+1, v);        getSons(root->right, d+1, v);        v[d].push_back(root->val);    }};

101. Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

    1   / \  2   2 / \ / \3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

    1   / \  2   2   \   \   3    3

Note:
Bonus points if you could solve it both recursively and iteratively.

Code:

/** * 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:    bool isSymmetric(TreeNode* root) {        if(root==NULL) return true;        return isSym(root->left, root->right);    }private:    bool isSym(TreeNode* l, TreeNode* r){        if(l==NULL && r==NULL) return true;                if(l!=NULL && r!=NULL){            if(l->val != r->val) return false;            return isSym(l->left, r->right) && isSym(l->right, r->left);        }                return false;    }};