[LeetCode]102. Binary Tree Level Order Traversal--二叉树层序遍历1

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]
]

---

解法1：利用队列先进先出的特性，先将父节点压入队列内，并依次检查其左右节点，非空则入队列，另外用两个变量记录本层需要遍历的节点数和下一层需要遍历的节点数即可。

/** * 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>> result;        if(root == NULL)            return result;        queue<TreeNode*> q;        q.push(root);                   // root不为空，压入队列        int toBePrint = 1;              // 记录本层还需要打印的节点个数        int nextLevel = 0;              // 记录下一层的节点个数，每次入队列加1        vector<int> oneLayer;           // 单层节点        while(!q.empty()){            TreeNode* pCur = q.front();            q.pop();            oneLayer.push_back(pCur->val);            if(pCur->left){             // 左节点不为空，入队列                q.push(pCur->left);                ++nextLevel;            }            if(pCur->right){            // 右结点不为空，入队列                q.push(pCur->right);                ++nextLevel;            }            if(--toBePrint == 0){       // toBePrint 为0，则将oneLayer压入result                toBePrint = nextLevel;                nextLevel = 0;                result.push_back(oneLayer);                oneLayer.clear();       // oneLayer清空            }        }        return result;    }};

解法2：利用两个队列分别存储本层节点和下一层结点

/** * 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>> result;        queue<TreeNode*> current, next;        if(root == NULL)             return result;        else             current.push(root);        while (!current.empty()) {            vector<int> oneLayer;             while (!current.empty()) {                TreeNode* pCur = current.front();                current.pop();                oneLayer.push_back(pCur->val);                if (pCur->left)                     next.push(pCur->left);                if (pCur->right)                     next.push(pCur->right);            }            result.push_back(oneLayer);            swap(next, current);              // 交换两层节点        }        return result;    }};

解法3：递归

/** * 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>> result;          traverse(root, 1, result);          return result;    }    void traverse(TreeNode* root, int level, vector<vector<int>>& result) {          if (root == NULL) return;          if (level > result.size())              result.push_back(vector<int>());          result[level-1].push_back(root->val);          traverse(root->left, level+1, result);          traverse(root->right, level+1, result);    }};