[Leetcode] Binary Tree Level Order Traversal & Binary Tree Zigzag Level Order Traversal
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这是两道题:
102. Binary Tree Level Order Traversal
103. Binary Tree Zigzag Level Order Traversal
102这道题就是按层次遍历,但是要把每一层的节点分别存储到vector里面去。
两年前就做过这道题,还写了博客。
当时的做法是硬生生地给每个结点加了个域,用来标记高度。很蹩脚。
今天做103的时候,回头看102,发现论坛里有几个巧妙的做法,所以悄悄地记录下。103等会儿讲。
首先是这位大神的做法。用的是队列,先把root和一个NULL指针放进去。然后依次取头结点,取到NULL的时候,就说明已经遍历完了一层,而且已经把这一层所有结点的子结点放入了队列,然后这个时候再放一个NULL进队列,就把这一层子结点和下一层给分开了。
class Solution {public: vector<vector<int> > levelOrder(TreeNode *root) { vector<vector<int> > result; if (!root) return result; queue<TreeNode*> q; q.push(root); q.push(NULL); vector<int> cur_vec; while(!q.empty()) { TreeNode* t = q.front(); q.pop(); if (t==NULL) { result.push_back(cur_vec); cur_vec.resize(0); if (q.size() > 0) { q.push(NULL); } } else { cur_vec.push_back(t->val); if (t->left) q.push(t->left); if (t->right) q.push(t->right); } } return result; }};
还有一位老哥的做法很简洁,用的是递归,先序遍历。堪称脑洞大开。
这样的做法就不用像我那样蹩脚地加个变量记录深度了。
class Solution {public:vector<vector<int> > levelOrder(TreeNode* root) {vector<vector<int> > ans;pre_order(root, 0, ans);return ans;}private:void pre_order(TreeNode* root, int depth, vector<vector<int> >& ans) {if (root == NULL)return;if (ans.size() == depth)ans.push_back(vector<int>());ans[depth].push_back(root->val);pre_order(root->left, depth + 1, ans);pre_order(root->right, depth + 1, ans);}};
然后将103这道题。其实也是按层次遍历返回vector,只不过规定第一层从左到右,第二层从右到左,第三层从左到右...
按照上面这位老哥的做法,写的代码:
class Solution {public:vector<vector<int> > zigzagLevelOrder(TreeNode* root) {vector<vector<int> > ans;pre_order(root, 0, ans);return ans;}private:void pre_order(TreeNode* root, int depth, vector<vector<int> >& ans) {if (root == NULL)return;if (ans.size() == depth)ans.push_back(vector<int>());if (depth % 2 == 0)ans[depth].push_back(root->val);elseans[depth].insert(ans[depth].begin(), root->val);pre_order(root->left, depth + 1, ans);pre_order(root->right, depth + 1, ans);}};
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