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1

10月24号面的fb。面试官是湾区白人小哥，出题难度中等。

第一题求binary tree上每个node的总和。
第二道给一个binary tree，要求打出垂直方向上的node。

花了一段时间才写出来，被提醒了几次，感觉有点磕磕绊绊。第二天收到拒信。
大家还是要好好[刷题]啊。

跪求~

2

啥是垂直方向上的node啊？

3

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

  1. Binary Tree Vertical Order Traversal

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 .

Input: [3,9,20,null,null,15,7]

   3
  /\
 /  \
 9  20
    /\
   /  \
  15   7 

Output:

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

Input: [3,9,8,4,0,1,7]

     3
    /\
   /  \
   9   8
  /\  /\
 /  \/  \
 4  01   7 

Output:

[
  [4],
  [9],
  [3,0,1],
  [8],
  [7]
]

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]
]
1 个赞
4

谢啦

5

The following solution takes 5ms .

  • BFS, put node , col into queue at the same time
  • Every left child access col - 1 while right child col + 1
  • This maps node into different col buckets
  • Get col boundary min and max on the fly
  • Retrieve result from cols

Note that TreeMap version takes 9ms .

Here is an example of [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15] . Notice that every child access changes one column bucket id. So 12 actually goes ahead of 11 .

image
image.jpg656×644 67.9 KB 

public List<List<Integer>> verticalOrder(TreeNode root) {
    List<List<Integer>> res = new ArrayList<>();
    if (root == null) {
        return res;
    }
    
    Map<Integer, ArrayList<Integer>> map = new HashMap<>();
    Queue<TreeNode> q = new LinkedList<>();
    Queue<Integer> cols = new LinkedList<>();

    q.add(root); 
    cols.add(0);

    int min = 0;
    int max = 0;
    
    while (!q.isEmpty()) {
        TreeNode node = q.poll();
        int col = cols.poll();
        
        if (!map.containsKey(col)) {
            map.put(col, new ArrayList<Integer>());
        }
        map.get(col).add(node.val);

        if (node.left != null) {
            q.add(node.left); 
            cols.add(col - 1);
            min = Math.min(min, col - 1);
        }
        
        if (node.right != null) {
            q.add(node.right);
            cols.add(col + 1);
            max = Math.max(max, col + 1);
        }
    }

    for (int i = min; i <= max; i++) {
        res.add(map.get(i));
    }

    return res;
}
3 个赞
6

Alternatively, we can calculate the rang first, then insert into buckets.

public List<List<Integer>> verticalOrder(TreeNode root) {
    List<List<Integer>> cols = new ArrayList<>();
    if (root == null) {
        return cols;
    }
    
    int[] range = new int[] {0, 0};
    getRange(root, range, 0);
    
    for (int i = range[0]; i <= range[1]; i++) {
        cols.add(new ArrayList<Integer>());
    }
    
    Queue<TreeNode> queue = new LinkedList<>();
    Queue<Integer> colQueue = new LinkedList<>();
    
    queue.add(root);
    colQueue.add(-range[0]);
    
    while (!queue.isEmpty()) {
        TreeNode node = queue.poll();
        int col = colQueue.poll();
        
        cols.get(col).add(node.val);
        
        if (node.left != null) {
            queue.add(node.left);   
            colQueue.add(col - 1);
        } 
        if (node.right != null) {
            queue.add(node.right);
            colQueue.add(col + 1);
        }
    }
    
    return cols;
}

public void getRange(TreeNode root, int[] range, int col) {
    if (root == null) {
        return;
    }
    range[0] = Math.min(range[0], col);
    range[1] = Math.max(range[1], col);
    
    getRange(root.left, range, col - 1);
    getRange(root.right, range, col + 1);
}
1 个赞
7

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