[C++]LeetCode: 29 Maximum Depth of Binary Tree

题目：

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.

递归的方法:

Anwser 1 ： DFS

思路：返回左子树或者右子树中大的深度加1，作为自己的深度即可

AC Code:

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int maxDepth(TreeNode *root) {        //计算二叉树的深度        //递归的方法        return root ? 1 + max(maxDepth(root->left), maxDepth(root->right)) : 0;    }};

非递归的方法:

Anwser 2：

BFS   in  queue

思路：用广度优先遍历的方法，队列先进先出，当一行元素pop结束时depth加1.

Attention:　队列queue指针front,　rear；注意队列的构造，调用，指针。

AC Code:

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int maxDepth(TreeNode *root) {            if(root == NULL) return 0;            queue<TreeNode*>  TreeQ;            TreeQ.push(root);      int cnt = 1;      int depth = 0;            while(!TreeQ.empty())      {          //队列queue指针front,rear          TreeNode* tmp = TreeQ.front();          TreeQ.pop();          cnt--;                    //如果结点左右孩子非空，则push          if(tmp->left)          {              TreeQ.push(tmp->left);          }          if(tmp->right)          {              TreeQ.push(tmp->right);          }                    if(cnt == 0)          {              depth++;               //一行结束，深度加1  // one row is end, add a depth                cnt = TreeQ.size();    //下一行结点的个数   // next row count          }      }            return depth;    }};