【leetcode】104. Maximum Depth of Binary Tree【java】三种实现方法：递归、BFS、DFS

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.

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

 //方法一：递归public class Solution {    public int maxDepth(TreeNode root) {         if (root == null){            return 0;        }        return 1 + Math.max(maxDepth(root.left), maxDepth(root.right));    }}//方法二 BFS 使用队列  这种方法相对快一些，也好一些public class Solution {    public int maxDepth(TreeNode root) {        if (root == null){            return 0;        }        Queue<TreeNode> queue = new LinkedList<TreeNode>();        queue.add(root);        int count = 0;        while (!queue.isEmpty()){            int size = queue.size();            for (int i = size; i > 0; i--){                TreeNode node = queue.poll();                if (node.left != null){                    queue.add(node.left);                }                if (node.right != null){                    queue.add(node.right);                }            }            count++;        }        return count;    }}//方法3 DFS 使用栈public class Solution {    public int maxDepth(TreeNode root) {        if (root == null){            return 0;        }        Stack<TreeNode> stack = new Stack<>();        Stack<Integer> value = new Stack<>();        int max = 0;        stack.push (root);        value.push(1);        while (!stack.isEmpty()){            TreeNode node = stack.pop();            int temp = value.pop();            max = Math.max(temp, max);            if (node.left != null){                stack.push(node.left);                value.push(temp + 1);            }            if (node.right != null){                stack.push(node.right);                value.push(temp + 1);            }        }        return max;    }}