Leetcode bfs&dfs Binary Tree Postorder Traversal II

Binary Tree Level Order Traversal II

  Total Accepted: 16983  Total Submissions: 54229 My Submissions

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]
]

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

题意：从底往上按层遍历二叉树
思路：
思路和Binary Tree Level Order Traveral 一样，
即从上往下按层遍历二叉树，将每一层的节点存放到该层对应的数组中
最后将得到的数组倒转一下就可以了
按层遍历二叉树可用bfs，也可用dfs，但都要记录节点所在的层
复杂度：时间O(n)， 空间O(n)

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void levelOrderBottom(TreeNode *root, int level, vector<vector<int> >&result)
    {
    	if(!root) return ;
    	if(level >= result.size()) { 
    		vector<int> temp;
    		temp.push_back(root->val);
    		result.push_back(temp);
    	}
    	else result[level].push_back(root->val);
    	levelOrderBottom(root->left, level + 1, result);
    	levelOrderBottom(root->right, level + 1, result);
    }
    
    vector<vector<int> > levelOrderBottom(TreeNode *root){
    	vector<vector<int> >result;
    	levelOrderBottom(root, 0, result);
    	reverse(result.begin(), result.end());
    	return result;
    }
};