leetcode【104】Maximum Depth of Binary Tree

问题描述：

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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.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its depth = 3.

源码：

先还是看看递归的做法吧，我自己写的非常复杂，太菜了。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int depth(TreeNode* node, int cur){
        if(!node->left && !node->right)     return cur;
        int tmp = 0;
        if(node->left)  tmp = depth(node->left, cur+1);
        if(node->right) tmp = max(tmp, depth(node->right, cur+1));
        return tmp;
    }
    int maxDepth(TreeNode* root) {
        if(!root)   return 0;
        return depth(root, 1);
    }
};

看看Discuss区大佬的做法，简单明了。

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

非递归方法，效率高了很多。时间87%，空间100%

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
/**
 * Definition for a binary tree node.
 * 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 *> q;
        q.push(root);
        int depth=0;// 记录深度
        while(!q.empty())
        {
            int len = q.size();//记录树当前层的结点个数
            ++depth;
            while(len)
            {
                TreeNode * p=q.front();
                if(p->left)
                    q.push(p->left);
                if(p->right)
                    q.push(p->right);
                q.pop();
                --len;
            }  
        }
        return depth;
    }
        
};