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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; } };
Leetcode:maximum_depth_of_binary_tree题解
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