Day7. Vertical_Order_Traversal_of_a_Binary_Tree.cpp

/*

Vertical Order Traversal of a Binary Tree

Given a binary tree, return the vertical order traversal of its nodes values.

For each node at position (X, Y), its left and right children respectively will be at positions (X-1, Y-1) and (X+1, Y-1).

Running a vertical line from X = -infinity to X = +infinity, whenever the vertical line touches some nodes, we report the values of the nodes in order from top to bottom (decreasing Y coordinates).

If two nodes have the same position, then the value of the node that is reported first is the value that is smaller.

Return an list of non-empty reports in order of X coordinate.  Every report will have a list of values of nodes.

 

Example 1:



Input: [3,9,20,null,null,15,7]
Output: [[9],[3,15],[20],[7]]
Explanation: 
Without loss of generality, we can assume the root node is at position (0, 0):
Then, the node with value 9 occurs at position (-1, -1);
The nodes with values 3 and 15 occur at positions (0, 0) and (0, -2);
The node with value 20 occurs at position (1, -1);
The node with value 7 occurs at position (2, -2).
Example 2:



Input: [1,2,3,4,5,6,7]
Output: [[4],[2],[1,5,6],[3],[7]]
Explanation: 
The node with value 5 and the node with value 6 have the same position according to the given scheme.
However, in the report "[1,5,6]", the node value of 5 comes first since 5 is smaller than 6.
 

Note:

1. The tree will have between 1 and 1000 nodes.
2. Each node's value will be between 0 and 1000.

*/

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
   
    vector<vector<int>> verticalTraversal(TreeNode* root) 
    {
        unordered_map<int,vector<pair<int,int>>> hash;
        int mn=INT_MAX;
        int mx=INT_MIN;
        vector<vector<int>> sol;
        if(root==nullptr) return sol;
        
        int x=0;
        queue<pair<TreeNode*,int>> q;
        q.push({root,0});
        int y=0;
        while(!q.empty())
        {
            int n=q.size();
            while(n--)
            {
                auto t=q.front();
                q.pop();
                hash[t.second].push_back({y,t.first->val});
                mn=min(mn,t.second);
                mx=max(mx,t.second);
                if(t.first->left)
                    q.push({t.first->left,t.second-1});
                if(t.first->right)
                    q.push({t.first->right,t.second+1});
            }
            y++;
            
        }
        
        for(int i=mn;i<=mx;i++)
        {
            sort(hash[i].begin(),hash[i].end());
            vector<int> curr;
            for(auto x : hash[i])
            {
                curr.push_back(x.second);
            }
            sol.push_back(curr);
        }
        return sol;
    }
};