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,#,#,15,7},

    3
   /   9  20
    /     15   7

return its bottom-up level order traversal as:

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

使用BFS层序遍历，然后得到的vector数组使用reverse反转即可。

Code(c++):

/**
 * 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;
        if(!root) return res;
        vector<int> nums;
        queue<TreeNode *> que;

        TreeNode *p = root;
        que.push(p);

        while(!que.empty()){ 
            int queSize = que.size();
            nums.resize(0);
            for(int i = 0; i < queSize; i++) {
                p = que.front();
                nums.push_back(p->val);
                if(p->left) que.push(p->left);
                if(p->right) que.push(p->right);
                que.pop();
            }
            res.push_back(nums);
        }
        reverse(res.begin(),res.end());
        return res;

    }
};