102:Binary Tree Level Order Traversal【树】【BFS】

题目链接：click~

解法1：

/*题意：将二叉树中的结点一层一层的输出*/

/**
 *思路：用两个队列，一个队列保存当前层的结点，另一队列保存下一层的结点。
 *      当遍历某层结点时，插入下一层的结点。
 */

class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<int> v;
        vector<vector<int>> path;

        if(root == NULL) return path;

        queue<TreeNode *> curr;
        curr.push(root);
        while(!curr.empty()) {
            queue<TreeNode *> next;//下一层结点
            while(!curr.empty()) {
                TreeNode *p = curr.front();
                v.push_back(p->val);
                curr.pop();

                //将下一层结点插入
                if(p->left != NULL) next.push(p->left);
                if(p->right != NULL) next.push(p->right);
            }
            curr = next;
            path.push_back(v);
            v.clear();
        }
        return path;
    }
};

解法2：

class Solution {
public:
    void DFS(TreeNode *T, int level, vector<vector<int>> &path) {
        if(T == NULL) return;
        if(level == path.size()) {
            path.resize(level+1);
        }
        path[level].push_back(T->val);
        DFS(T->left, level+1, path);
        DFS(T->right, level+1, path);
    }
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int>> path;
        if(root == NULL) return path;
        DFS(root, 0, path);
    }
};