题目链接 : https://leetcode-cn.com/problems/maximum-depth-of-binary-tree/

题目描述:

给定一个二叉树，找出其最大深度。

二叉树的深度为根节点到最远叶子节点的最长路径上的节点数。

说明: 叶子节点是指没有子节点的节点。

示例：
给定二叉树 [3,9,20,null,null,15,7]，

    3
   / \
  9  20
    /  \
   15   7
返回它的最大深度 3 。

思路:

思路一:DFS

思路二:BFS

代码:

思路一:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if not root:return 0
        return 1 + max(self.maxDepth(root.left), self.maxDepth(root.right))

java

/**
 * 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;
        return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));  
    }
}

思路二:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        from collections import deque
        if not root: return 0
        queue = deque()
        queue.appendleft(root)
        res = 0
        while queue:
            #print(queue)
            res += 1
            n = len(queue)
            for _ in range(n):
                tmp = queue.pop()
                if tmp.left:
                    queue.appendleft(tmp.left)
                if tmp.right:
                    queue.appendleft(tmp.right)
        return res

java

/**
 * 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;
        Deque<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        int res = 0;
        while (!queue.isEmpty()) {
            res++;
            int n = queue.size();
            for (int i = 0; i < n; i++) {
                TreeNode tmp = queue.poll();
                if (tmp.left != null) queue.add(tmp.left);
                if (tmp.right != null) queue.add(tmp.right);
            }
        }
        return res;
    }
}