lintcode,二叉树的最大深度

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距离。

解题思路：递归解乏就是判断左右子树最大深度哪一个更大。还可以用两个栈分别存储节点和节点所在深度，只要节点栈不为空，就取出深度判断是否更大，然后将左右孩子节点放进节点栈和节点深度栈。
一刷ac。二刷ac

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: An integer.
     */
    public int maxDepth(TreeNode root) {
        if(root == null) return 0;
        return Math.max(maxDepth(root.left), maxDepth(root.right))+1;
    }
}

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: An integer.
     */
    public int maxDepth(TreeNode root) {
        if(root == null) return 0;
        Stack nodestack = new Stack();
        Stack depthstack = new Stack();
        int maxdepth = 1;
        nodestack.push(root);
        depthstack.push(maxdepth);
        while(!nodestack.isEmpty()){
            TreeNode node = nodestack.pop();
            int d = depthstack.pop();
            if(d > maxdepth) maxdepth = d;
            if(node.left != null){
                nodestack.push(node.left);
                depthstack.push(d+1);
            }
            if(node.right != null){
                nodestack.push(node.right);
                depthstack.push(d+1);
            }
        }
        return maxdepth;
    }
}