314. 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 1:

Input: [3,9,20,null,null,15,7]

   3
  /\
 /  \
 9  20
    /\
   /  \
  15   7 

Output:

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

Examples 2:

Input: [3,9,8,4,0,1,7]

     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7 

Output:

[
  [4],
  [9],
  [3,0,1],
  [8],
  [7]
]

Examples 3:

Input: [3,9,8,4,0,1,7,null,null,null,2,5] (0’s right child is 2 and 1’s left child is 5)

     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7
    /\
   /  \
   5   2

Output:

[
  [4],
  [9,5],
  [3,0,1],
  [8,2],
  [7]
]

方法1:

用BFS遍历，并将高度Hd 和 Node * 存入hash table
https://www.youtube.com/watch?v=PQKkr036wRc

易错点：

1. 注意queue里面是pair<int, TreeNode*>, 要用myQueue.front().first 和myQueue.front().second来获取
2. 这里用dfs遍历会有一个顺序问题：高层的按照要求应该先放入但是这个层级顺序在dfs中不好掌握，除非用双map存一下x，y坐标。

class Solution {
public:
    vector<vector<int>> verticalOrder(TreeNode* root) {
        if (!root) return {};
        map<int, vector<int>> hash;
        queue<pair<TreeNode*, int>> q;
        q.push(make_pair(root, 0));
        while (!q.empty()) {
            TreeNode* top = q.front().first;
            int pos = q.front().second;
            q.pop();
        
            hash[pos].push_back(top -> val);
            
            if (top -> left) q.push(make_pair(top -> left, pos - 1));
            if (top -> right) q.push(make_pair(top -> right, pos + 1));
        }
        vector<vector<int>> res;
        for (auto item: hash) res.push_back(item.second);
        return res;
    }
};