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Leetcode 树 Binary Tree Level Order Traversal II

2024-07-04 10:22:10   226人阅读

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Binary Tree Level Order Traversal II

 Total Accepted: 10080 Total Submissions: 32610

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)

相关题目：

Binary Tree Level Order Traversal 

Binary Tree Zigzag Level Order Traversal


/**
 * 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;
    }


};


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