987. Vertical Order Traversal of a Binary Tree

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DFS

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class Solution {
    public List<List<Integer>> verticalTraversal(TreeNode root) {
        Map<Integer, Queue<int[]>> colIndexToNumListMap = new TreeMap<>(); // 值选择优先队列而不是 TreeSet 是因为可能存在重复元素
        dfs(colIndexToNumListMap, root, 0, 0);
        List<List<Integer>> resultList = new ArrayList<>();
        for (Queue<int[]> numQueue : colIndexToNumListMap.values()) {
            List<Integer> numList = new ArrayList<>(numQueue.size());
            while (!numQueue.isEmpty()) { // 注意优先队列整体无序，使用 for 循环遍历并非有序，所以此处依次弹出最小值节点
                numList.add(numQueue.poll()[2]);
            }
            resultList.add(numList);
        }
        return resultList;
    }

    private void dfs(Map<Integer, Queue<int[]>> colIndexToNumListMap, TreeNode node, int i, int j) {
        if (node == null) {
            return;
        }

        Queue<int[]> queue = colIndexToNumListMap.computeIfAbsent(j, k -> new PriorityQueue<>((o1, o2) -> {
            int diff = Integer.compare(o1[0], o2[0]);
            if (diff != 0) {
                return diff;
            } else {
                return Integer.compare(o1[2], o2[2]);
            }
        }));
        queue.offer(new int[]{i, j, node.val});

        dfs(colIndexToNumListMap, node.left, i + 1, j - 1);
        dfs(colIndexToNumListMap, node.right, i + 1, j + 1);
    }
}

以上解法通过在 DFS 遍历过程中构建结果集。

DFS

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class Solution {
    private static class TreeNodeWrapper {
        private final int val;
        private final int row;
        private final int col;

        public TreeNodeWrapper(int val, int row, int col) {
            this.val = val;
            this.row = row;
            this.col = col;
        }
    }

    public List<List<Integer>> verticalTraversal(TreeNode root) {
        List<List<Integer>> resultList = new ArrayList<>();

        List<TreeNodeWrapper> treeNodeWrapperList = new ArrayList<>();
        dfs(treeNodeWrapperList, root, 0, 0);

        Collections.sort(treeNodeWrapperList, (o1, o2) -> {
            if (o1.col != o2.col) {
                return Integer.compare(o1.col, o2.col);
            } else {
                if (o1.row != o2.row) {
                    return Integer.compare(o1.row, o2.row);
                } else {
                    return Integer.compare(o1.val, o2.val);
                }
            }
        });

        for (int i = 0; i < treeNodeWrapperList.size(); i++) {
            TreeNodeWrapper curr = treeNodeWrapperList.get(i);
            if (i == 0) {
                List<Integer> list = new ArrayList<>();
                list.add(curr.val);
                resultList.add(list);
            } else {
                TreeNodeWrapper prev = treeNodeWrapperList.get(i - 1);
                if (curr.col == prev.col) {
                    resultList.get(resultList.size() - 1).add(curr.val);
                } else {
                    List<Integer> list = new ArrayList<>();
                    list.add(curr.val);
                    resultList.add(list);
                }
            }
        }

        return resultList;
    }

    private void dfs(List<TreeNodeWrapper> treeNodeWrapperList, TreeNode node, int row, int col) {
        if (node == null) {
            return;
        }

        treeNodeWrapperList.add(new TreeNodeWrapper(node.val, row, col));
        dfs(treeNodeWrapperList, node.left, row + 1, col - 1);
        dfs(treeNodeWrapperList, node.right, row + 1, col + 1);
    }
}

以上解法通过 DFS 收集所有节点至一个平级列表，再排序拆分为题目要求的二级列表。

References

987. Vertical Order Traversal of a Binary Tree
PriorityQueue (Java Platform SE 8)
Sorting Priority Queue in Java
The built-in iterator for java’s PriorityQueue does not traverse the data structure in any particular order. Why?