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.
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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 ;
q.push(root);
int res = ;
while(!q.empty()){
res++;
for(int i = q.size(); i > ; 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 ;
return + 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 = ;
int maxDepth(TreeNode* root) {
int depth = ;
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+);
DFS(root->right,depth+);
}
};