987. Vertical Order Traversal of a Binary Tree

最新推荐文章于 2021-03-11 16:53:46 发布

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本文介绍了一种使用TreeMap和优先队列解决LeetCode上二叉树垂直遍历问题的方法。通过递归深度优先搜索，将节点位置信息映射到TreeMap中，并利用优先队列确保同一位置上的节点按值从小到大排列。

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.
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题目链接

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

方法思路

TreeMap 是一个有序的key-value集合，它是通过红黑树实现的。
TreeMap 继承于AbstractMap，所以它是一个Map，即一个key-value集合。

优先队列的作用是能保证每次取出的元素都是队列中权值最小的（Java的优先队列每次取最小元素，C++的优先队列每次取最大元素）。
Java中PriorityQueue通过二叉小顶堆实现，可以用一棵完全二叉树表示。

    //Runtime: 3 ms, faster than 98.30% 
    //Memory Usage: 37.5 MB, less than 22.46%
    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()) {
            list.add(new ArrayList<>());
            for (PriorityQueue<Integer> nodes : ys.values()) {
                while (!nodes.isEmpty()) {
                    list.get(list.size() - 1).add(nodes.poll());
                }
            }
        }
        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);
    }
}