[LeetCode107]Binary Tree Level Order Traversal II 二叉树层次遍历

题目：

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]
]

分类：Tree BFS 

代码：

递归：

class Solution {
protected:
    vector<vector<int>> ans;
    void dfs(TreeNode *root, int height){
        if (root == NULL) 
            return;
        while (ans.size() <= height)
            ans.push_back(vector<int>());
        ans[height].push_back(root->val);
        dfs(root->left, height + 1);
        dfs(root->right, height + 1);
    }

public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
       dfs(root, 0);
       reverse(ans.begin(), ans.end());
        return ans;    
    }
};

非递归：

/**
 * 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) {
        vector<vector<int>> resultVec;
        if(!root)
            return resultVec;
        int parentSize = 1, childSize = 0;
        int level = 0;
        TreeNode *temp;
        queue<TreeNode*> nodes;
        nodes.push(root);
        resultVec.push_back(vector<int>());
        while(!nodes.empty())
        {
            temp  = nodes.front();
            resultVec[level].push_back(temp->val);
            nodes.pop();
            if(temp->left)
            {
                nodes.push(temp->left);
                childSize++;
            }
            if(temp->right)
            {
                nodes.push(temp->right);
                childSize++;
            }
            parentSize--;
            if(parentSize == 0)
            {
                parentSize = childSize;
                childSize = 0;
                level++;
                resultVec.push_back(vector<int>());
            }
        }
        resultVec.erase(resultVec.end()-1);
        reverse(resultVec.begin(),resultVec.end());
        return resultVec;
    }
};