[LeetCode314]Binary Tree Vertical Order Traversal

Given a binary tree, return the vertical order traversal of its nodes' values. (ie, from top to bottom, column by column).

If two nodes are in the same row and column, the order should be from left to right.

Examples:
Given binary tree [3,9,20,null,null,15,7],
        3
       / \
      9  20
      /   \
     15   7
return its vertical order traversal as:
    [
      [9],
      [3,15],
      [20],
      [7]
    ]
Given binary tree [3,9,20,4,5,2,7],
    _3_
   /   \
  9    20
 / \   / \
4   5 2   7
return its vertical order traversal as:
    [
      [4],
      [9],
      [3,5,2],
      [20],
      [7]
    ]
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這題一眼就可以看出： 對於current root， 它和 root->left->right and root->left->right 在一個vector 裏。我看出這個以後就再也看不出別的了（苦笑。。

看了別人的實現方法，很好玩～BFS + hashmap。
如果current root 的index 是idx: let the index of left node be idx-1 and index of right node be idx+1 then the index of root->left->right is also idx!!!(same with root->right->left)

/**
 * 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>> verticalOrder(TreeNode* root) {
        map<int, vector<int>> mp;
        queue<pair<int, TreeNode*>> q;
        vector<vector<int>> res;
        if(!root) return res;
        q.push(make_pair(0,root));
        while(!q.empty()){
            int sz = q.size();
            for(int i = 0; i<sz; ++i){
                TreeNode* cur = q.front().second;
                int idx = q.front().first;
                q.pop();
                mp[idx].push_back(cur->val);
                if(cur->left) q.push(make_pair(idx-1, cur->left));
                if(cur->right) q.push(make_pair(idx+1,cur->right));
            }
        }
        for(auto m : mp){
            res.push_back(m.second);
        }
        return res;
    }
};