(二叉树 BFS DFS) leetcode 104. Maximum Depth of Binary Tree

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 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);
    }
};