107. Binary Tree Level Order Traversal II
Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree [3,9,20,null,null,15,7]
,
3 / \ 9 20 / \ 15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
解题思路:记录每行的节点数,now表示当前层的节点数,pre表示上一层的节点数,每次出队的的时候让pre--,直到减为0,将now赋值给pre,同时让now继续为0.
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class Solution { public: vector<vector<int>> levelOrderBottom(TreeNode* root) { if(!root)return {}; vector<vector<int>>ans; vector<int>level; queue<TreeNode* >q; q.push(root); int now=0,pre=1; while(!q.empty()){ root=q.front(); level.push_back(root->val); q.pop(); pre--; if(root->left)q.push(root->left),now++; if(root->right)q.push(root->right),now++; if(pre==0){ ans.push_back(level); level.clear(); pre=now; now=0; } } reverse(ans.begin(),ans.end()); return ans; } };