LeetCode—树—层次遍历
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2022-03-13 12:58:59
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LeetCode—树—层次遍历
使用 BFS 进行层次遍历。不需要使用两个队列来分别存储当前层的节点和下一层的节点,因为在开始遍历一层的节点时,当前队列中的节点数就是当前层的节点数,只要控制遍历这么多节点数,就能保证这次遍历的都是当前层的节点。
1、一棵树每层节点的平均数
T637. Average of Levels in Binary Tree (Easy)
/**
* 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<double> averageOfLevels(TreeNode* root) {
if(!root) return {};
vector<double> res;
queue<TreeNode*> q{{root}};
while(!q.empty()){
int n=q.size();
double sum=0;
for(int i=0; i<n; ++i){
TreeNode* t = q.front();
q.pop();
sum += t->val;
if(t->left) q.push(t->left);
if(t->right) q.push(t->right);
}
res.push_back(sum/n);
}
return res;
}
};
2、得到左下角的节点
T513. Find Bottom Left Tree Value (Easy)
/**
* 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:
int findBottomLeftValue(TreeNode* root) {
int res=0;
queue<TreeNode*> q{{root}};
while(!q.empty()){
int n=q.size();
for(int i=0; i<n; ++i){
TreeNode* t = q.front();
q.pop();
if(i == 0) res = t->val;
if(t->left) q.push(t->left);
if(t->right) q.push(t->right);
}
}
return res;
}
};