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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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LeetCode-Binary Tree Level Order Traversal II