LeetCode—树—层次遍历

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;


    }
};