LeetCode 104 Maximum Depth of Binary Tree 二叉树最大深度

LeetCode 104 Maximum Depth of Binary Tree
难度：Easy

题目描述：
找到一颗二叉树的最深深度。
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.

解题思路：
Solution 1：
解二叉树的题目一般如果难度为easy，则要求iterative和recursive都会写，二叉树的最深深度是左子树和右子树的Max深度，根据这一特性，我们自底向上，时间复杂度O(n), Space O(1), 但因为是递归，所以会占用stack，recursive的写法如下：

public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        //get the left and right depth
        int leftDep = maxDepth(root.left);
        int rightDep = maxDepth(root.right);
        return 1 + Math.max(leftDep, rightDep);
    }

解题思路：
Solution 2：
使用BFS方法iterative使用一个Queue，每往下一层都加1，最后拿到最深深度。
时间：O(n), 空间 O(n)

public int maxDepth(TreeNode root) {
        //iterative
        if (root == null) return 0;
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        int dep = 0;
        while(!q.isEmpty()) {
            dep++;
            int size = q.size();
            for(int i = 0; i < size; i++) {
                TreeNode now = q.poll();
                if (now.left != null) q.offer(now.left);
                if (now.right != null) q.offer(now.right);
            }

        }
        return dep;
        
    }