Binary Tree Vertical Order Traversal

https://www.lintcode.com/problem/binary-tree-vertical-order-traversal/description

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

 3

/\

/ \

9 8

/ /\

/ \/ \

4 01 7

return its vertical order traversal as:

[

[4],

[9],

[3,0,1],

[8],

[7]

]

Thoughts

纵向遍历二叉树，结点最多的那层最左结点为起始层。把root所在纵向坐标看作0，其它结点存与它的相对坐标，即左叶子为parent - 1, 右+1，然后按层遍历。

Code

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param root: the root of tree
     * @return: the vertical order traversal
     */
    vector<vector<int>> verticalOrder(TreeNode * root) {
        if (root == nullptr) return vector<vector<int>>();
        queue<pair<TreeNode*, int>> q;
        q.push({root, 0});
        unordered_map<int, vector<int>> m;
        int mi = 0, mx = 0;
        while (!q.empty()) {
            const auto t = q.front();
            q.pop();
            m[t.second].push_back(t.first->val);
            mi = min(mi, t.second);
            mx = max(mx, t.second);
            if (t.first->left != nullptr) q.push({t.first->left, t.second - 1});
            if (t.first->right != nullptr) q.push({t.first->right, t.second + 1});
        }
        vector<vector<int>> res;
        for (int i = mi; i <= mx; ++i) {
            res.push_back(m[i]);
        }
        return res;
    }
};

Analysis

时间复杂度O(N), N为结点数.