[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,#,#,15,7},

    3   / \  9  20    /  \   15   7

return its bottom-up level order traversal as:

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

跟之前的题目一样，就是最后加一个反转。

recursive solution

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

iterative solution

/** * 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>> result;        if(!root)   return result;                queue<TreeNode*> current, next;        vector<int> level;        current.push(root);        while(!current.empty())        {            while(!current.empty())            {                TreeNode *node = current.front();                current.pop();                level.push_back(node->val);                if(node->left)  next.push(node->left);                if(node->right) next.push(node->right);            }            result.push_back(level);            level.clear();            swap(next, current);        }        reverse(result.begin(), result.end());        return result;    }};