LeetCode - Binary Tree Level Order Traversal II

题目描述：

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).

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]]

从上到下得到的层序遍历反过来即可。

class Solution {public:    vector<vector<int>> levelOrderBottom(TreeNode* root)     {        queue<pair<TreeNode*,int> > q; vector<pair<int,int> > v;if(root==NULL) {    vector<vector<int> > result;    return result;}else if(root!=NULL){pair<TreeNode*,int> p;p.first=root;p.second=0;q.push(p);while(!q.empty()){p=q.front();pair<int,int> x;x.first=p.first->val;x.second=p.second;v.push_back(x);q.pop();if(p.first->left!=NULL){pair<TreeNode*,int> l;l.first=p.first->left;l.second=p.second+1;q.push(l);}if(p.first->right!=NULL){pair<TreeNode*,int> r;r.first=p.first->right;r.second=p.second+1;q.push(r);}}}int n=v.size();int m=v[n-1].second;vector<vector<int> > result(m+1);for(int i=0;i<n;i++){result[v[i].second].push_back(v[i].first);}vector<vector<int> > Result(m+1);for(int i=0,j=m;i<=m&&j>=0;i++,j--){    Result[i]=result[j];}return Result;    }};