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 7return its depth = 3.
-----------------------------------------------------------------------------------------------------------------------------
求二叉树的最大深度。
emmm，用bfs时，注意要用循环，要先遍历完一层，再遍历下一层。和leetcode111 Minimum Depth of Binary Tree几乎相似，只是少写了一行代码而已。
C++代码：
/** * 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) { queue<TreeNode*> q; if(!root) return 0; q.push(root); int res = 0; while(!q.empty()){ res++; for(int i = q.size(); i > 0; i--){ //必须写循环，否则在[3,9,20,null,null,15,7]时，会返回5。嗯，就是返回了二叉树的结点的个数。 auto t = q.front(); q.pop(); if(t->left) q.push(t->left); if(t->right) q.push(t->right); } } return res; } };
也可以用DFS，如果能够理解递归，就能够很好的理解DFS了。
C++代码：
/** * 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) return 0; return 1 + max(maxDepth(root->left),maxDepth(root->right)); } };还有一个递归，是自顶向下的递归
C++代码：
/** * 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 answer = 0; int maxDepth(TreeNode* root) { int depth = 1; DFS(root,depth); return answer; } void DFS(TreeNode* root,int depth){ if(!root) return; if(!root->left && !root->right) answer = max(answer,depth); DFS(root->left,depth+1); DFS(root->right,depth+1); } };
转载于:https://www.cnblogs.com/Weixu-Liu/p/10714904.html
相关资源：JAVA上百实例源码以及开源项目