Given a binary tree, return thebottom-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 as:
[ [15,7], [9,20], [3]]
/** * 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) {queue<TreeNode*> nodes;vector< vector<int> > result;vector<int> tmp;if(NULL==root)return result;nodes.push(root);while(!nodes.empty()){int length=nodes.size();int i=0;while(i<length){TreeNode* tmpNode=nodes.front();tmp.push_back(tmpNode->val);if(tmpNode->left)nodes.push(tmpNode->left);if(tmpNode->right)nodes.push(tmpNode->right);nodes.pop();i++;}result.push_back(tmp);tmp.clear();}reverse(result.begin(),result.end());return result;}};又是层序遍历的方式，层序遍历反复出现过多次：求二叉树的最小深度、右侧能看到的结点、统计特殊形式的二叉树结点个数…
，真正的寂寞是在人群中，当你面对许多熟悉的脸，
