二叉树的最小深度

[题目](111. 二叉树的最小深度)

重点

使用前序遍历

代码实现

/**
 * 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 result;
    void getdepth(TreeNode* node, int depth) {
        if (node->left == NULL && node->right == NULL) {
            result = min(result, depth);
            return;
        }
        // 中 只不过中没有处理的逻辑
        if (node->left) { // 左
            getdepth(node->left, depth + 1);
        }
        if (node->right) { // 右
            getdepth(node->right, depth + 1);
        }
    }

    int minDepth(TreeNode* root) {
        if (root == NULL) {
            return 0;
        }
        result = INT_MAX;
        getdepth(root, 1);
        return result;
    }
};

完全二叉树的节点个数

[题目](222. 完全二叉树的节点个数)

重点

主要是求出深度

代码实现

/**
 * 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 == NULL) {
            return 0;
        }
        TreeNode* left = root->left;
        TreeNode* right = root->right;
        int leftDepth = 0, rightDepth = 0;
        // 求左子树深度
        while (left) {
            left = left->left;
            leftDepth++;
        }
        // 求右子树深度
        while (right) {
            right = right->right;
            rightDepth++;
        }
        if (leftDepth == rightDepth) {
            // 2 << 1 相当于 2^2
            return (2 << leftDepth) - 1;
        }
        return countNodes(root->left) + countNodes(root->right) + 1;
    }
};