Maximum Depth of Binary Tree

题目104：Maximum Depth of Binary Tree

题目描述：
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.
思路分析：
和求二叉树的最小深度类似的解法类似，有递归和非递归两种解法。

1. 递归解法

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

2. 非递归解法

思路分析：
二叉树的层次遍历，一直遍历到最后一个结点，此时的深度就是求的深度。

class Solution {public:    int maxDepth(TreeNode* root) {        if (0 == root)            return 0;        queue<TreeNode *> que;        int depth, i, len;        que.push(root);        depth = 1;        while (!que.empty()) {            len = que.size();            for (i = 0; i < len; i ++) {                TreeNode *qtop = que.front();                que.pop();                if (qtop->left)                    que.push(qtop->left);                if (qtop->right)                    que.push(qtop->right);            }            if (que.empty())                return depth;            ++ depth;        }    }};

while里面有for循环，还要优化的地方。
用cuc_num统计当前层的结点个数，next_num统计下一层的结点个数。

class Solution {public:    int maxDepth(TreeNode* root) {        if (0 == root)            return 0;        queue<TreeNode *> que;        int cur_num, next_num, depth;        que.push(root);        depth = 1;        cur_num = 1;        next_num = 0;        while (!que.empty()) {            TreeNode *qtop = que.front();            que.pop();            -- cur_num;            if (qtop->left) {                que.push(qtop->left);                ++ next_num;            }            if (qtop->right) {                que.push(qtop->right);                ++ next_num;            }            /* 判断队列是否为空，如果为空，深度不再需要加1 */            if (!que.empty() && 0 == cur_num) {                cur_num = next_num;                next_num = 0;                ++ depth;            }        }        return depth;    }};