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,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7]
[9,20],
[3],
]
Solution: Queue version. On the basis of 'Binary Tree Level Order Traversal', reverse the final vector.
NULL is used to make gap between levels.
1 /** 2 * Definition for binary tree 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> > res; 14 queue<TreeNode*> q; 15 if(!root) return res; 16 q.push(root); 17 q.push(NULL); 18 vector<int> level; 19 20 while(true) { 21 TreeNode* node = q.front(); 22 q.pop(); 23 if(node) { 24 level.push_back(node->val); 25 if(node->left) q.push(node->left); 26 if(node->right) q.push(node->right); 27 } 28 else { 29 res.push_back(level); 30 level.clear(); 31 if(q.empty()) break; 32 q.push(NULL); 33 } 34 } 35 36 reverse(res.begin(), res.end()); 37 return res; 38 } 39 };
