#70 Binary Tree Level Order Traversal II

題目描述：

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).

Have you met this question in a real interview? 

Yes

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]
]

題目思路：

這題和#71基本一樣，只是在BFS之後，把得到的ans reverse一下即可（因爲要求bottom to top）.

Mycode（AC = 21ms）：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
 
 
class Solution {
    /**
     * @param root : The root of binary tree.
     * @return : buttom-up level order a list of lists of integer
     */
public:
    vector<vector<int>> levelOrderBottom(TreeNode *root) {
        // write your code here
        vector<vector<int>> ans;
        if (root == NULL) return ans;
        
        // initialize 2 queues, one to 
        // store tree nodes, one to store
        // level/height of corresponding node
        queue<TreeNode *> helper;
        helper.push(root);
        queue<int> levels;
        levels.push(0);
        
        while (!helper.empty()) {
            TreeNode *node = helper.front();
            helper.pop();
            
            int level = levels.front();
            levels.pop();
            
            // push the node into ans, according
            // to its level
            if (ans.size() <= level) {
                ans.push_back({node->val});
            }
            else {
                ans[level].push_back(node->val);
            }
            
            // push the left/right tree and their
            // corresponding level into queue
            if (node->left) {
                helper.push(node->left);
                levels.push(level + 1);
            }
            if (node->right) {
                helper.push(node->right);
                levels.push(level + 1);
            }
        }
        
        // reverse the final ans, because it requires bottom to top
        reverse(ans.begin(), ans.end());
        return ans;
    }
};