Leetcode:maximum_depth_of_binary_tree题解

一、     题目

   给定一个二叉树，求它的最大深度。最大深度是沿从根节点，到叶节点最长的路径。

二、     分析

   (做到这里发现接连几道题都是用递归，可能就是因为自己时挑的简单的做的吧。)

   找出最深路径，则每经过一个节点要加上1，当遇到空节点时，返回0。

   在网上看了看有的做法除了麻烦点但是很不错的方法，比如使用栈和队列等

/** * 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;        return max(maxDepth(root->left),maxDepth(root->right))+1;    }};队列/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    //二叉树最大深度（层次遍历，遍历一层高度加1）    int maxDepth(TreeNode *root) {        int height = 0,rowCount = 1;        if(root == NULL){            return 0;        }        //创建队列        queue<treenode*> queue;        //添加根节点        queue.push(root);        //层次遍历        while(!queue.empty()){            //队列头元素            TreeNode *node = queue.front();            //出队列            queue.pop();            //一层的元素个数减1，一层遍历完高度加1            rowCount --;            if(node->left){                queue.push(node->left);            }            if(node->right){                queue.push(node->right);            }            //一层遍历完            if(rowCount == 0){                //高度加1                height++;                //下一层元素个数                rowCount = queue.size();            }        }        return height;    } };</treenode*> 栈/** * 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;                     stack<treenode*> S;                     int maxDepth = 0;          TreeNode *prev = NULL;                     S.push(root);          while (!S.empty()) {              TreeNode *curr = S.top();                             if (prev == NULL || prev->left == curr || prev->right == curr) {                  if (curr->left)                      S.push(curr->left);                  else if (curr->right)                      S.push(curr->right);              } else if (curr->left == prev) {                  if (curr->right)                      S.push(curr->right);              } else {                  S.pop();              }              prev = curr;              if (S.size() > maxDepth)                  maxDepth = S.size();          }          return maxDepth;      }  };</treenode*> DFS/** * 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;        int res = INT_MIN;        dfs(root, 1, res);        return res;    }    void dfs(TreeNode *root, int depth, int &res)    {        if(root->left == NULL && root->right == NULL && res < depth)            {res = depth; return;}        if(root->left)            dfs(root->left, depth+1, res);        if(root->right)            dfs(root->right, depth+1, res);    }};层次遍历/** * 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) {        //层次遍历树的层数,NULL为每一层节点的分割标志        if(root == NULL)return 0;        int res = 0;        queue<TreeNode*> Q;        Q.push(root);        Q.push(NULL);        while(Q.empty() == false)        {            TreeNode *p = Q.front();            Q.pop();            if(p != NULL)            {                if(p->left)Q.push(p->left);                if(p->right)Q.push(p->right);            }            else             {                res++;                if(Q.empty() == false)Q.push(NULL);            }        }        return res;    }};