Binary Tree Level Order Traversal II -- leetcode

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]
]

 基本思路：

同问题 Binary Tree Level Order Traversal 一样，

在结果返回前，作一个reverse操作。

/**
 * 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> > ans;
        dfs(ans, root, 0);
        reverse(ans.begin(), ans.end());
        return ans;
    }
    
    void dfs(vector<vector<int> > &ans, TreeNode *root, int level) {
        if (!root) return;
        if (level == ans.size())
            ans.push_back(vector<int>());
        
        ans[level].push_back(root->val);
        dfs(ans, root->left, level+1);
        dfs(ans, root->right, level+1);
    }
};

Binary Tree Level Order Traversal II -- leetcode

原文：http://blog.csdn.net/elton_xiao/article/details/45581679