LeetCode Binary Tree Level Order Traversal II

class Solution {
private:
    vector<vector<int> > nodes;
public:
    vector<vector<int> > levelOrderBottom(TreeNode *root) {
        nodes.clear();
        dfs(root, 0);
        reverse(nodes.begin(), nodes.end());
        return nodes;
    }
    void dfs(TreeNode* root, int level) {
        if (root == NULL) return;
        if (level >= nodes.size()) {
            nodes.push_back(vector<int>());
        }
        nodes[level].push_back(root->val);
        dfs(root->left, level + 1);
        dfs(root->right, level + 1);
    }
};

复杂度应该不会有不同， 应该有更好的方法，不用做reverse

第二轮：

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]
]

几乎一样的代码，可以用链表避免做reverse


第三轮：
先求出树德最大高度就可以避免做reverse。

/**
 * 7:13
 * 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 {
private:
    vector<vector<int> > res;
public:
    vector<vector<int>> levelOrderBottom(TreeNode* root) {
        int h = height(root);
        res = vector<vector<int> >(h);
        if (h == 0) {
            return res;
        }
        dfs(root, h - 1);
        return res;
    }
    
    int height(TreeNode* root) {
        if (root == NULL) {
            return 0;
        }
        int dl = height(root->left);
        int dr = height(root->right);
        
        return 1 + max(dl, dr);
    }
    
    void dfs(TreeNode* root, int idx) {
        if (root == NULL) {
            return; 
        }
        
        dfs(root->left, idx - 1);
        dfs(root->right, idx - 1);
        
        res[idx].push_back(root->val);
    }
};