代码随想录算法训练营第十六天| 104.二叉树的最大深度、111.二叉树的最小深度、222.完全二叉树的节点个数

104.二叉树的最大深度

题目链接
文章讲解

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode* root) {
        int ans = 0;
        queue<TreeNode*> que;
        TreeNode* node;
        if (root)
            que.push(root);
        while (!que.empty()) {
            int size = que.size();
            while (size--) {
                node = que.front();
                que.pop();
                if (node->left)
                    que.push(node->left);
                if (node->right)
                    que.push(node->right);
            }
            ans++;
        }
        return ans;
    }
};

111.二叉树的最小深度

题目链接
文章讲解

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        queue<TreeNode*> que;
        TreeNode* node;
        int depth = 0;
        if (root)
            que.push(root);
        while (!que.empty()) {
            depth++;
            int size = que.size();
            while (size--) {
                node = que.front();
                que.pop();
                if (!node->left && !node->right)
                    return depth;
                if (node->left)
                   que.push(node->left);
                if (node->right)
                   que.push(node->right);
            }
        }
        return depth;
    }
};

222.完全二叉树的节点个数

题目链接
文章讲解

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int countNodes(TreeNode* root) {
        if (!root)
         	return 0;
        TreeNode* left = root->left;
        TreeNode* right = root->right;
        int leftDepth = 1, rightDepth = 1;
        while (left)
            left = left->left, leftDepth++;
        while (right)
            right = right->right, rightDepth++;
        if (leftDepth == rightDepth)
            return (1 << leftDepth) - 1;
        return countNodes(root->left) + countNodes(root->right) + 1;
    }
};