一、题目内容

image.png

二、题解

解法1：

思路

深度优先；求左子树与右子树最大深度，然后+1

代码

  1. /**
  2.  * Definition for a binary tree node.
  3.  * public class TreeNode {
  4.  *     int val;
  5.  *     TreeNode left;
  6.  *     TreeNode right;
  7.  *     TreeNode(int x) { val = x; }
  8.  * }
  9.  */
  10. class Solution {
  12.     /**
  13.      * 递归
  14.      *
  15.      * @param root
  16.      * @return
  17.      */
  18.     public int maxDepth(TreeNode root) {
  19.         if (root == null) {
  20.             return 0;
  21.         }
  23.         return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));
  24.     }
  25. }

解法2：

思路

广度优先；层序遍历，每遍历一层，则计数器 +1+1

代码

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {

    /**
     * 广度优先，层序遍历
     *
     * @param root
     * @return
     */
    public int maxDepth(TreeNode root) {
        int depth = 0;
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        if (root != null) {
            queue.offer(root);
        }
        while (!queue.isEmpty()) {
            int levelSize = queue.size();
            for (int i = 0; i < levelSize; i++) {
                TreeNode head = queue.poll();
                if (head.left != null) {
                    queue.offer(head.left);
                }
                if (head.right != null) {
                    queue.offer(head.right);
                }
            }
            depth++;
        }
        return depth;
    }
}