Binary Tree Level Order Traversal

这几个题目相似。都是用同样的递归法,时间空间复杂度都为O（n).

当然，还有迭代法，控件复杂度会降至 O(1），这个暂不讨论。

Binary Tree Level Order Traversal

// 递归版,时间复杂度 O(n),空间复杂度 O(n) class Solution {
public:
      vector<vector<int> > levelOrder(TreeNode *root) {
          vector<vector<int>> result;
          traverse(root, 1, result);
          return result;
}
      void traverse(TreeNode *root, size_t level, vector<vector<int>> &result) {
          if (!root) 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);
} };

Binary Tree Level Order Traversal II

// 递归版,时间复杂度 O(n),空间复杂度 O(n)
class Solution {
public:
vector<vector<int> > levelOrderBottom(TreeNode *root) { vector<vector<int>> result;
traverse(root, 1, result);
std::reverse(result.begin(), result.end()); // 比上一题多此一行 return result;
}
      void traverse(TreeNode *root, size_t level, vector<vector<int>> &result) {
          if (!root) 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);
} };

Binary Tree Zigzag Level Order Traversal

// 递归版,时间复杂度 O(n),空间复杂度 O(n)
class Solution {
public:
      vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
          vector<vector<int>> result;
          traverse(root, 1, result, true);
          return result;
}
      void traverse(TreeNode *root, size_t level, vector<vector<int>> &result,
              bool left_to_right) {
          if (!root) return;
          if (level > result.size())
              result.push_back(vector<int>());
          if (left_to_right)
              result[level-1].push_back(root->val);
          else
              result[level-1].insert(result[level-1].begin(), root->val);
          traverse(root->left, level+1, result, !left_to_right);
          traverse(root->right, level+1, result, !left_to_right);
      }
};