leetcode 之Binary Tree Level Order Traversal II-优快云博客

本文介绍了一种解决二叉树底部向上层次遍历的方法，通过先进行从顶部到底部的层次遍历（BFS），再反转结果来实现。文章提供了一个具体的C++实现示例。

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题目：

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},

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.

解答：

先从高到低遍历，然后reverse一下就可以了，就是一个BFS

/**
* 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>> res;
if(!root)
return res;
queue<TreeNode*> q;
q.push(root);
while(!q.empty())
{
vector<int> temp;
int size = q.size();
while(size--)
{
temp.push_back(q.front()->val);
if(q.front()->left != NULL)
q.push(q.front()->left);
if(q.front()->right != NULL)
q.push(q.front()->right);
q.pop();
}
res.push_back(temp);
}
reverse(res.begin(),res.end());
return res;
}
};