【LeetCode从零单刷】Maximum Depth of Binary Tree

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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);        }    }};