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# 987.Vertical-Order-Traversal-of-a-Binary-Tree

## 题目描述

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: [,[3,15],,]
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: [,,[1,5,6],,]
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:
The tree will have between 1 and 1000 nodes.
Each node's value will be between 0 and 1000.

## 代码

### Approach #2 Nested TreeMap + PriorityQueue => (x, y, root.val)

class Solution {
public List<List<Integer>> verticalTraversal(TreeNode root) {
TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map = new TreeMap<>();
dfs(root, 0, 0, map);
List<List<Integer>> list = new ArrayList<>();
for (TreeMap<Integer, PriorityQueue<Integer>> ys : map.values()) {
for (PriorityQueue<Integer> nodes : ys.values()) {
while (!nodes.isEmpty()) {
}
}
}
return list;
}
private void dfs(TreeNode root, int x, int y, TreeMap<Integer, TreeMap<Integer, PriorityQueue<Integer>>> map) {
if (root == null) return;
if (!map.containsKey(x)) {
map.put(x, new TreeMap<>());
}
if (!map.get(x).containsKey(y)) {
map.get(x).put(y, new PriorityQueue<>());
}
map.get(x).get(y).offer(root.val);
dfs(root.left, x - 1, y + 1, map);
dfs(root.right, x + 1, y + 1, map);
}
}

### Approach #1 Store Locations

class Solution {
List<Location> locations;
public List<List<Integer>> verticalTraversal(TreeNode root) {
// Each location is a node's x position, y position, and value
locations = new ArrayList();
dfs(root, 0, 0);
Collections.sort(locations);
List<List<Integer>> ans = new ArrayList();
int prev = locations.get(0).x;
for (Location loc: locations) {
// If the x value changed, it's part of a new report.
if (loc.x != prev) {
prev = loc.x;
}
// We always add the node's value to the latest report.
}
return ans;
}
public void dfs(TreeNode node, int x, int y) {
if (node != null) {
dfs(node.left, x-1, y+1);
dfs(node.right, x+1, y+1);
}
}
}
class Location implements Comparable<Location>{
int x, y, val;
Location(int x, int y, int val) {
this.x = x;
this.y = y;
this.val = val;
}
@Override
public int compareTo(Location that) {
if (this.x != that.x)
return Integer.compare(this.x, that.x);
else if (this.y != that.y)
return Integer.compare(this.y, that.y);
else
return Integer.compare(this.val, that.val);
}
}