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]
]

思路

和题目102. Binary Tree Level Order Traversal思路一致，对结果进行排序处理即可：vector<vector<int>>(res.rbegin(), res.rend());

c++11

/**
 * 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:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
        BFS(root, 0, res);
        return vector<vector<int>>(res.rbegin(), res.rend());
    }

    void BFS(TreeNode* root, int level, vector<vector<int>> &res) {
        if(!root) return;
        if(res.size() == level) res.push_back({});
        res[level].push_back(root->val);
        if(root->left) BFS(root->left, level+1, res);
        if(root->right) BFS(root->right, level+1, res);
    }
};