637. Average of Levels in Binary Tree 104. Maximum Depth of Binary Tree

Given a non-empty binary tree, return the average value of the nodes on each level in the form of an array.

Example 1:

Input:
    3
   / \
  9  20
    /  \
   15   7
Output: [3, 14.5, 11]
Explanation:
The average value of nodes on level 0 is 3,  on level 1 is 14.5, and on level 2 is 11. Hence return [3, 14.5, 11].

Note:

1. The range of node's value is in the range of 32-bit signed integer.

我的解答：

/**
 * 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) {
        vector<double> v;
        if(!root){
            return v;
        }
        
        queue<TreeNode*> q;
        q.push(root);
        
        while(!q.empty()){
            int s = q.size();
            long temp = 0;
            for(int i = 0; i < s; ++i){
                TreeNode* tp = q.front();
                q.pop();
                temp += (long)tp->val;
                if(tp->left)q.push(tp->left);
                if(tp->right)q.push(tp->right);
            }
            v.push_back(temp/(double)s);
        }
        return v;
    }
};

利用BFS，二叉樹的層次遍歷。

中間利用一個for循環來達到區分每一層的目的，很巧妙，標記一下。

第104，求一棵二叉數的最深的深度，有種解法，DFS和BFS

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node

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:
    int maxDepth(TreeNode* root) {
        int depth = 0;
        if(!root){
            return 0;
        }
        
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty()){
            depth++;
            int s = q.size();
            for(int i = 0; i < s; ++i){
                TreeNode* temp = q.front();
                q.pop();
                if(temp->left)q.push(temp->left);
                if(temp->right)q.push(temp->right);
            }

        }
        return depth;
    }
};

其中需要注意的地方是在while循環中的那個for循環，i < s不能寫爲i < q.size(),因爲在循環中q的size()變化了。

//DFS，利用遞歸,代碼很簡潔。
int maxDepth(TreeNode *root)
{
    return root == NULL ? 0 : max(maxDepth(root -> left), maxDepth(root -> right)) + 1;
}

.