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.

例子

例子 1

Input: [3,9,20,null,null,15,7] Output: [[9],[3,15],[20],[7]] **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).

例子 2

Input: [1,2,3,4,5,6,7] Output: [[4],[2],[1,5,6],[3],[7]] **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.

解题思路

可以根据题目的思路，对树节点进行（按行列）编号。遍历一次树将所有树节点进行编号，放入一个容器中，这里我用的是 map 可以顺便把编号的排序也做了，最后只需要对每一行每一列的向量进行排序即可（或者用set也可以），代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> verticalTraversal(TreeNode* root) {
        std::vector<std::vector<int>> results;

        std::map<int, std::map<int, std::vector<int>>> node_map;
        dfs(root, 0, 0, node_map);
        for (auto iter_col = node_map.begin(); iter_col != node_map.end(); iter_col++) {
            std::vector<int> col;
            std::map<int, std::vector<int>> col_map = iter_col->second;
            for (auto iter_row = col_map.begin(); iter_row != col_map.end(); iter_row++) {
                std::sort(iter_row->second.begin(), iter_row->second.end());
                col.insert(col.end(), iter_row->second.begin(), iter_row->second.end());
            }
            results.push_back(col);
        }

        return results;
    }

private:
    void dfs(TreeNode* node, int row, int col, std::map<int, std::map<int, std::vector<int>>>& node_map) {
        if (!node) return;
        node_map[col][row].push_back(node->val);
        dfs(node->left, row + 1, col - 1, node_map);
        dfs(node->right, row + 1, col + 1, node_map);
    }
};

• 时间复杂度: O(nlogn) （最坏情况是在分支只有一条的情况，不太确定）
• 空间复杂度: O(n)