C++二叉树叶结点个数

计算二叉树叶子结点个数可以通过递归的方式进行。如果当前节点为空，则返回0；如果当前节点的左右子节点都为空，则说明当前节点为叶子节点，返回1；否则，递归计算左子树和右子树的叶子节点个数，并将它们相加返回。

以下是C++的实现：

#include<iostream>
using namespace std;

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

int countLeafNodes(TreeNode* root) {
    if (root == NULL) {
        return 0;
    }
    if (root->left == NULL && root->right == NULL) {
        return 1;
    }
    return countLeafNodes(root->left) + countLeafNodes(root->right);
}

int main() {
    TreeNode *root = new TreeNode(1);
    root->left = new TreeNode(2);
    root->right = new TreeNode(3);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(5);
    root->right->left = new TreeNode(6);
    root->right->right = new TreeNode(7);

    int count = countLeafNodes(root);
    cout << "Leaf nodes count: " << count << endl;

    return 0;
}

输出结果为：

Leaf nodes count: 4