給定一個二叉樹，找出其最大深度。

二叉樹的深度為根節點到最遠葉子節點的最長路徑上的節點數。

說明: 葉子節點是指沒有子節點的節點。

示例：
給定二叉樹 [3,9,20,null,null,15,7]，

    3
   / \
  9  20
    /  \
   15   7

返回它的最大深度 3 。

方法一：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        return root == null? 0:(1+Math.max(maxDepth(root.left),maxDepth(root.right)));
    }
}
 

/**
 * 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));
    }
};
 

方法二：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null) return 0;
        int res = 0;
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        while(!q.isEmpty()){
            ++res;
            int n = q.size();
            for(int i = 0; i < n; i++){
                TreeNode t = q.poll();
                if(t.left != null) q.offer(t.left);
                if(t.right != null) q.offer(t.right);
            }
        }
        return res;
    }
}
 

/**
 * 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;
        int res = 0;
        queue<TreeNode *> q;
        q.push(root);
        while(!q.empty()){
            res++;
            int n = q.size();
            for(int i = 0; i < n; i++){
                TreeNode *t = q.front();
                q.pop();
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
        }
        return res;
    }
};