314. Binary Tree Vertical Order Traversal

• Difficulty: Medium

• Topics: Hash Table

• Similar Questions:

  • Binary Tree Level Order Traversal

Problem:

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]
]

Solutions:

/**
 * 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) {
        queue<pair<TreeNode*, int>> q;
        map<int, vector<int>> verticals;

        q.push({root,0});

        while(!q.empty()) {
            auto node = q.front(); q.pop();
            if (node.first == nullptr) continue;
            verticals[node.second].push_back(node.first->val);
            q.push({node.first->left, node.second - 1});
            q.push({node.first->right, node.second + 1});
        }
        vector<vector<int>> ret;
        for (auto it = verticals.begin(); it != verticals.end(); ++it) {
            ret.push_back(it->second);
        }

        return ret;

    }
};