Leetcode 107. Binary Tree Level Order Traversal II
107. Binary Tree Level Order Traversal II
Total Accepted: 85871 Total Submissions: 247487 Difficulty: Easy
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] ]
思路：实质是求自底向上的层序情况。可以见该题的姐妹题题解：Leetcode 102. Binary Tree Level Order Traversal
代码：
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 vector<vector<int>> levelOrderBottom(TreeNode* root) { 13 vector<vector<int>> v; 14 if(root==NULL) return v; 15 queue<TreeNode*> q; 16 q.push(root); 17 TreeNode *begin,*end=root,*cur=root; 18 while(!q.empty()){ 19 vector<int> temp; 20 begin=q.front(); 21 q.pop(); 22 while(begin!=end){ 23 temp.push_back(begin->val); 24 if(begin->left){ 25 q.push(begin->left); 26 cur=begin->left; 27 } 28 if(begin->right){ 29 q.push(begin->right); 30 cur=begin->right; 31 } 32 begin=q.front(); 33 q.pop(); 34 } 35 temp.push_back(begin->val); 36 if(begin->left){ 37 q.push(begin->left); 38 cur=begin->left; 39 } 40 if(begin->right){ 41 q.push(begin->right); 42 cur=begin->right; 43 } 44 v.insert(v.begin(),temp); //相较于102题，就是这里变动了 45 end=cur; 46 } 47 return v; 48 } 49 };
