[Leetcode] Binary Tree level Order traversal II

Binary Tree level Order traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes ' values. (ie, from the left-to-right, the level by level from the leaf to root).

For example:
Given binary Tree {3,9,20,#,#,15,7} ,

    3   /   9    /   7

Return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]]

Confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

Problem Solving Ideas:

The problem test instructions is to traverse the binary tree from bottom to top, and the result exists in a two-dimensional vector. You can find the height of the tree, determine the size of the first dimension of the two-dimensional vector, and then depth through the enumeration to insert the value into the corresponding position. The following method requires only 8ms. Another problem given in the topic: {1,#,2,3} means storing a binary tree as an array, #表示对应的节点为空, subscript can represent the relationship of two nodes.

/** * 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) {int h = getHeight (r        OOT);        vector<vector<int>> result (h, vector<int> ());        Traversal (root, 1, h, result);    return result; }private:void traversal (treenode* root, int level, int h, vector<vector<int>>& result) {if (root            !=null) {result[h-level].push_back (root->val);            Traversal (Root->left, level+1, h, result);        Traversal (Root->right, level+1, h, result);        }} int getheight (treenode* root) {if (root==null) {return 0;    } return 1+max (GetHeight (Root->left), GetHeight (Root->right)); }};

[Leetcode] Binary Tree level Order traversal II