树 求树的最小深度、最大深度

一、二叉树的最小深度

Given a binary tree, find its minimum depth.The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

求解此类问题要先明确递归状态及递归条件，考虑清楚是否有漏洞。本题有两种解法。

解法一：分别存储左右子树的深度，返回他们中的较小值。注意：要考虑到只有一个子节点的情况。

Class Solution {

public:
    int run(TreeNode *root) {
        if(!root) return 0;
        int l = 0, r = 0;
        if(root->left)  l = run(root->left);
        if(root->right)  r = run(root->right);
        if(root->left && root->right) return min(l, r)+1;
        else return l+r+1;

    }
};

解法二：

Class Solution {

public:
    int run(TreeNode *root) {
        if(!root) return 0;
        if(!root->left)  return run(root->right)+1;
        if(!root->right)  return run(root->left)+1;
        return min(run(root->right), run(root->left))+1;       
    }
};

二、二叉树的最大深度

链接：https://www.nowcoder.com/questionTerminal/8a2b2bf6c19b4f23a9bdb9b233eefa73
来源：牛客网

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.

同上述最小深度方法类似，也可用两种方法求解：

class Solution {
public:
    int maxDepth(TreeNode *root) {
        if(!root) return 0;
        int l = 0, r = 0;
        if(root->right) r = maxDepth(root->right);
        if(root->left) l = maxDepth(root->left);
        return max(l, r)+1;
    }
};

class Solution {
public:
    int maxDepth(TreeNode *root) {
        if(!root) return 0;
        if(root->right && root->left) 
            return max(maxDepth(root->left), maxDepth(root->right))+1;
        if(root->right) return maxDepth(root->right)+1;
        if(root->left) return maxDepth(root->left)+1;
        return 1;
    }
};