Binary Tree Level Order Traversal II问题及解法

问题描述：

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).

示例：

Given binary tree [3,9,20,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

问题分析：

要获取每一层的结点信息，首推树的层次遍历BFS。每遍历一层，就可以将它存储到数组中。

过程详见代码：

/** * 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> > res;        if(root == NULL) return res;        queue<TreeNode*> q,tq;        q.push(root);        tq.push(root);                while(!tq.empty())        {        vector<int> v;        while(!q.empty())        {        TreeNode* node = q.front();        v.push_back(node->val);        if(node->left != NULL)        tq.push(node->left);        if(node->right != NULL)        tq.push(node->right);        tq.pop();        q.pop();}res.insert(res.begin(),v);q = tq;}        return res;    }};

代码不难理解~~~