5.1.2 Binary Tree Level Order Traversal II

Notes: 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. */   /** * 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>> root2leaf;        queue<TreeNode *> q;        if (!root) return root2leaf;        q.push(root);        q.push(NULL);   // end indicator of one level        vector<int> level;        while (true)        {            TreeNode *node = q.front(); q.pop();            if (node)            {                level.push_back(node->val);                if (node->left) q.push(node->left);                if (node->right) q.push(node->right);            }            else            {                root2leaf.push_back(level);                level.clear();                if (q.empty()) break;    // CAUTIOUS! infinite loop                q.push(NULL);            }        }    // reverse        reverse(root2leaf.begin(), root2leaf.end());        return root2leaf;    }};