LeetCode-Binary Tree Level Order Traversal II

题目链接：https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

题目：

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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

解题思路：这道题就是Binary Tree Level Order Traversal的变化，改的不多，直接对矩阵进行倒置输出就行。

/** * Definition for binary tree * 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> > s;        vector<vector<int> > k;        vector<int> t;        int nowNum = 0;    //nowNum        int nextNum = 0;   //nextNum        if (root == NULL) return s;        queue<TreeNode *> q;        q.push(root);        nowNum++;        while(!q.empty()) {            TreeNode *node = q.front();            q.pop();            nowNum--;            t.push_back(node->val);            if (node->left != NULL) {                q.push(node->left);                nextNum++;            }            if (node->right != NULL) {                q.push(node->right);                nextNum++;            }            if (nowNum == 0) {                s.push_back(t);                nowNum = nextNum;                nextNum = 0;                t.clear();            }        }        int m = s.size();        for (int i = m-1; i>=0;i--)            k.push_back(s[i]);        return k;    }};

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