LeetCode--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：和之前Binary Tree Level Order Traversal这个一样，只不过最后reverse反转一下，倒序。

方法一：递归。

/** * 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>>result;        traverse(root,result,1);        reverse(result.begin(),result.end());        return result;    }    void traverse(TreeNode* root,vector<vector<int>>&result,int level){        if(!root) return;        if(level>result.size()){            result.push_back(vector<int>());        }        result[level-1].push_back(root->val);        traverse(root->left,result,level+1);        traverse(root->right,result,level+1);    }};

方法二：迭代。

/** * 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>>result;        traverse(root,result,1);        reverse(result.begin(),result.end());        return result;    }    void traverse(TreeNode* root,vector<vector<int>>&result,int level){        if(!root) return;        if(level>result.size()){            result.push_back(vector<int>());        }        result[level-1].push_back(root->val);        traverse(root->left,result,level+1);        traverse(root->right,result,level+1);    }};