Given a binary tree, returnBottom-up level orderTraversal 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:
[ [15,7], [9,20], [3]]
In fact, the bottom sequence traversal still starts from the top, but the storage method has changed. For more information, see my previous blog http://www.cnblogs.com/grandyang/p/4051321.html. The Code is as follows:
/** * Definition for binary tree * 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) { vector<vector<int> > res; if (root == NULL) return res; queue<TreeNode*> q; q.push(root); while (!q.empty()) { vector<int> oneLevel; int size = q.size(); for (int i = 0; i < size; ++i) { TreeNode *node = q.front(); q.pop(); oneLevel.push_back(node->val); if (node->left) q.push(node->left); if (node->right) q.push(node->right); } res.insert(res.begin(), oneLevel); } return res; }};
Binary Tree level order traversal II Binary Tree sequence traversal II