[Swift]LeetCode314. 二叉树的竖直遍历 $ Binary Tree Vertical Order Traversal

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➤微信公众号：山青咏芝（shanqingyongzhi）
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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:
Given binary tree [3,9,20,null,null,15,7],

    3
   / 
  9  20
    /  
   15   7

return its vertical order traversal as:

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

Given binary tree [3,9,20,4,5,2,7],

     3
   /   
  9    20
 /    / 
4   5 2   7

return its vertical order traversal as:

[
  [4],
  [9],
  [3,5,2],
  [20],
  [7]
]

---

给定二叉树，返回其节点值的垂直顺序遍历。（即自上而下逐列）。

如果两个节点在同一行和同一列中，则顺序应该从左到右。

实例：

给定的二叉树[3,9,20,null,null,15,7],

    3
   / 
  9  20
    /  
   15   7

返回其垂直顺序遍历为：

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

给定二叉树

[3,9,20,4,5,2,7],

     3
   /   
  9    20
 /    / 
4   5 2   7

返回其垂直顺序遍历为：

[
  [4],
  [9],
  [3,5,2],
  [20],
  [7]
]

---

Solution:

public class TreeNode {
    public var val: Int
    public var left: TreeNode?
    public var right: TreeNode?
    public init(_ val: Int) {
        self.val = val
        self.left = nil
        self.right = nil
    }
}

class Solution {
    func verticalOrder(_ root: TreeNode?) ->[[Int]] {
        var res:[[Int]] = [[Int]]()
        if root == nil {return res}
        var m:[Int:[Int]] = [Int:[Int]]()
        var q:[(Int,TreeNode?)] = [(Int,TreeNode?)]()
        q.append((0, root))
        while(!q.isEmpty)
        {
            var a:(Int,TreeNode?) = q.first!
            q.popLast()
            m[a.0,default:[Int]()].append(a.1!.val)
            if a.1?.left != nil
            {
                q.append((a.0 - 1, a.1!.left))
            }
            if a.1?.right != nil
            {
                q.append((a.0 + 1, a.1!.right))
            }
        }
        for a in m
        {
            res.append(a.1)
        }
        return res             
    }
}