LeetCode Maximum Depth of Binary Tree

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.

思路分析：这题比較简单。关于树的题目通常都能够用递归解决，这题也不例外。递归解法的思路是返回左右子树中深度较大的子树深度加1作为自己的深度。每递归调用一次深度加1.

递归解法（DFS递归搜索）

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

非递归的解法须要借助队列实现，BFS搜索树节点，首先root入队列。然后不断从队列头取出节点，将该节点的左右孩子（假设有）入队列。直到队列为空。注意维护当前层次的节点数和下一层次的节点数。这样在每次换层的时候，把层次计数器level加1，最后返回level层次数作为结果就可以。因为每一个node訪问一次。时间复杂度O（n）。

非递归解法（借助队列BFS）

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int maxDepth(TreeNode root) {
        if(root == null) return 0;
        LinkedList<TreeNode> treeNodeQueue = new LinkedList<TreeNode>();
        int level = 0;
        treeNodeQueue.add(root);
        int curLevelNum = 1;//# of nodes in the current level
        int nextLevelNum = 0;//# of nodes in the next level
        while(!treeNodeQueue.isEmpty()){
            TreeNode curNode = treeNodeQueue.poll();//find and remove; different from peek
            curLevelNum--;
            if(curNode.left != null){
                treeNodeQueue.add(curNode.left);
                nextLevelNum++;
            }
            if(curNode.right != null){
                treeNodeQueue.add(curNode.right);
                nextLevelNum++;
            }
            if(curLevelNum == 0){
                level++;
                curLevelNum = nextLevelNum;
                nextLevelNum = 0;//added a level
            }
        }
        return level;
    }
}