对一张琴
一壶酒
一溪云
C++，层序遍历，和上题几乎一样的。
思路直接看上一道题吧。
非递归：
/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
*/
class Solution {
public:
vector<vector<int> > Print(TreeNode* pRoot) {
vector<vector<int>> ans;
if (!pRoot) return ans;
queue<TreeNode *> q;
q.push(pRoot);
while (!q.empty()) {
vector<int> tempv;
int cnt = q.size();
while (cnt) {
TreeNode *cur = q.front();
q.pop();
tempv.push_back(cur->val);
if (cur->left)
q.push(cur->left);
if (cur->right)
q.push(cur->right);
cnt--;
}
ans.push_back(tempv);
}
return ans;
}
};
递归：
/*
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
*/
class Solution {
public:
vector<vector<int> > Print(TreeNode* pRoot) {
vector<vector<int>> ans;
dfs(pRoot, ans, 0);
return ans;
}
void dfs(TreeNode *root, vector<vector<int>> &ans, int depth) {
if (!root) return ;
if (depth >= ans.size())
ans.push_back(vector<int> {});
ans[depth].push_back(root->val);
dfs(root->left, ans, depth + 1);
dfs(root->right, ans, depth + 1);
}
};
