LeetCode 107. Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

]

这个题目要采用广度优先遍历，所以我们需要一个队列，为了记录节点的深度我用了一个map映射来记录节点的深度

当vec的长度小于节点的深度时，就需要插入一个新的向量，否则直接在向量上插入节点的val

最后我们再翻转vec，即为所求答案

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void breadthFirstSearch(TreeNode* root, vector<vector<int>> &vec)
    {
        if (!root)return;
        queue<TreeNode*> que;
        map<TreeNode*, int> ma;
        que.push(root);
        ma[root] = ;
        while (!que.empty())
        {
            TreeNode* node = que.front();
            if (node->left)
            {
                que.push(node->left);
                ma[node->left] = ma[node] + ;
            }
            if (node->right)
            {
                que.push(node->right);
                ma[node->right] = ma[node] + ;
            }
            if (vec.size() <= ma[node])
            {
                vector<int> childvec;
                childvec.push_back(node->val);
                vec.push_back(childvec);
            }
            else
            {
                vec[ma[node]].push_back(node->val);
            }
            que.pop();

        }
    }
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> vec;
        breadthFirstSearch(root, vec);
        reverse(vec.begin(), vec.end());
        return vec;
    }
};