【二叉树】107. Binary Tree Level Order Traversal II

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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

解答：

看起来没什么挑战性。。。就是把层次遍历反转。。

/** * 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) return res;        queue<TreeNode*> q;        q.push(root);        while(!q.empty()){            vector<int> tmp;            for(int i=0,n=q.size();i<n;i++){                TreeNode* p=q.front();                q.pop();                tmp.push_back(p->val);                if(p->left)                    q.push(p->left);                if(p->right)                    q.push(p->right);            }            res.push_back(tmp);        }        reverse(res.begin(),res.end());        return res;    }};

一遍AC，美滋滋