[LeetCode] 102. Binary Tree Level Order Traversal

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).
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]
]

解析

二叉树的层次遍历，有递归和非递归（利用队列queue）两种方法。

解法1：队列

每一层的节点个数，利用queue的size来得到，循环处理每一层。

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode* > q;
        vector<vector<int> > res;
        if(!root) return res;
        q.push(root);
        while(!q.empty()){
            int lengh = q.size();
            vector<int> v;
            while(lengh--){
                TreeNode* front = q.front();
                v.push_back(front->val);
                q.pop();
                if(front->left) q.push(front->left);
                if(front->right) q.push(front->right);
            }
            res.push_back(v);
        }
        return res;
    }
};

解法2：递归

用一个level变量来定义层，当层数和当前res大小相同时，需要增添一层，即res.push_back({})。

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int> > res;
        if(!root) return res;
        levelorder(root, 0, res);
        return res;
    }
    void levelorder(TreeNode* root, int level, vector<vector<int> > &res){
        if(!root) return;
        if (res.size() == level) res.push_back({});   // 层数与res大小相同时，增加一个空vector
        res[level].push_back(root->val);
        if(root->left) levelorder(root->left, level+1, res);
        if(root->right) levelorder(root->right, level+1, res);
    }
};

参考

http://www.cnblogs.com/grandyang/p/4051321.html