【LeetCode从零单刷】Maximum Depth of Binary Tree

没错我就是伍声2009的粉丝，从今天起，模仿《从零单排》系列，菜鸡单刷LeetCode！

题目：

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.

解答：

树的深度优先遍历，求得树高度即可。然后需要用到递归的思想，假设子树的高度已知。最后，记得设定递归终点。

说归说，我这样的菜鸡还是会碰到问题……例如下面这个：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode *root) {
        if(root->left == NULL && root->right == NULL)
        {
            return 1;
        }
        if(root->left != NULL && root->right == NULL)
        {
            return (1+maxDepth(root->left));
        }
        if(root->right != NULL && root->left == NULL)
        {
            return (1+maxDepth(root->right));
        }
        else
        {
            return (1+maxDepth(root->right))>(1+maxDepth(root->left))?(1+maxDepth(root->right)):(1+maxDepth(root->left));
        }
    }
};

Runtime Error 了……原因是，当输入空树时无法处理。但是改了之后还是还有通过……原因是超时。想了半天，问题在这里：

return (1+maxDepth(root->right))>(1+maxDepth(root->left))?(1+maxDepth(root->right)):(1+maxDepth(root->left));

这样不就是把子树的 maxDepth 求解了四次么？正确方法应该将子树高度保存下来，这样只需要计算两次，节省一半时间。得到 AC：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int maxDepth(TreeNode *root) {
        if(root == NULL)
        {
            return 0;
        }
        if(root->left == NULL && root->right == NULL)
        {
            return 1;
        }
        if(root->left != NULL && root->right == NULL)
        {
            return (1+maxDepth(root->left));
        }
        if(root->right != NULL && root->left == NULL)
        {
            return (1+maxDepth(root->right));
        }
        else
        {
            int l = maxDepth(root->left);
            int r = maxDepth(root->right);
            return (1+r)>(1+l)?(1+r):(1+l);
        }
    }
};