LeetCode 107 Binary Tree Level Order Traversal II题解

107.

Binary Tree Level Order Traversal II

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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]
]

解题思路：

本题与Binary Tree Level Order Traversal 方法相同，即用bfs进行层次遍历，最后使用reverse 函数将 遍历结果反向即可。

代码展示：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include <algorithm>
class Solution {
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        vector<vector<int>> res;
        if(!root) return res;
        queue<TreeNode*> tra;
        tra.push(root);
        //vector<int> v;
        //v.push_back(root->val);
        //res.push_back(v);
        int precount=1;
        int count =0;
        while(tra.size())
        {
            //cout<<tra.size()<<endl;
            vector<int> v;
            while(precount--)
            {
                TreeNode * top = tra.front();
                tra.pop();
                if(top->left)
                {
                    tra.push(top->left);
                    count++;
                }
                if(top->right)
                {
                    tra.push(top->right);
                    count++;
                }
                v.push_back(top->val);
                
            }
            if(v.size())
            res.push_back(v);
            precount = count;
            count=0;
            
            
        }
        reverse(res.begin(),res.end());
        return res;
    }
};