算法练习笔记（八）—— 树的BFS遍历

BFS遍历，广度优先搜索，在用法上来说和这周所学的贪心算法有着很大程度上的相似

与深度优先算法不同的是，算法会在遍历完一棵树中每一层的每个结点之后，才会转到下一层继续，在这一基础上，二维数组和队列将会对算法起到很大的帮助

题目地址：https://leetcode.com/problems/binary-tree-level-order-traversal/#/description

题目：Binary Tree Level Order Traversal

描述：

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:

Given binary tree 

[3,9,20,null,null,15,7]

,

3
/ \
9  20
/  \
15   7

return its level order traversal as:

[
[3],
[9,20],
[15,7]
]

BFS解答：

/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *
4000
left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {

vector<vector<int>> result;

public:
void BFS(TreeNode * root, int dep){
if(!root)return;
if(result.size() == dep)
result.push_back(vector<int>());
result[dep].push_back(root -> val);
BFS(root -> left, dep + 1);
BFS(root -> right, dep + 1);
}
vector<vector<int>> levelOrder(TreeNode* root) {
BFS(root, 0);
return result;
}
};