# 987. 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`.

``````class Solution {
public:
vector<vector<int>> verticalTraversal(TreeNode* root) {
vector<vector<int>> res;
map<int, map<int, set<int>>> m;
queue<pair<TreeNode*, vector<int>>> q;
q.push({root, {0, 0}});
while (!q.empty()) {
auto t = q.front(); q.pop();
TreeNode *node = t.first;
int x = t.second[0], y = t.second[1];
m[x][y].insert(node->val);
if (node->left) q.push({node->left, {x - 1, y + 1}});
if (node->right) q.push({node->right, {x + 1, y + 1}});
}
for (auto &a : m) {
vector<int> col;
for (auto &it : a.second) {
col.insert(col.end(), it.second.begin(), it.second.end());
}
res.push_back(col);
}
return res;
}
};
``````

``````class Solution {
public:
vector<vector<int>> verticalTraversal(TreeNode* root) {
vector<vector<int>> res;
map<int, map<int, set<int>>> m;
helper(root, 0, 0, m);
for (auto &a : m) {
vector<int> col;
for (auto &it : a.second) {
col.insert(col.end(), it.second.begin(), it.second.end());
}
res.push_back(col);
}
return res;
}
void helper(TreeNode* root, int x, int y, map<int, map<int, set<int>>>& m) {
if (!root) return;
m[x][y].insert(root->val);
helper(root->left, x - 1, y + 1, m);
helper(root->right, x + 1, y + 1, m);
}
};
``````

Github 同步地址:

https://github.com/grandyang/leetcode/issues/987

Binary Tree Vertical Order Traversal

https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree/

https://leetcode.com/problems/vertical-order-traversal-of-a-binary-tree/discuss/260502/C%2B%2B-BFSDFS