# 543.Diameter of Binary Tree

**543.Diameter of Binary Tree**

难度:Easy

> 给定一棵二叉树，你需要计算它的直径长度。一棵二叉树的直径长度是任意两个结点路径长度中的最大值。这条路径可能穿过根结点。

```
示例 :
给定二叉树

          1
         / \
        2   3
       / \     
      4   5    
返回 3, 它的长度是路径 [4,2,1,3] 或者 [5,2,1,3]。

注意：两结点之间的路径长度是以它们之间边的数目表示。
```

直观方法：最大直径是左子树深度和右子树深度之和，与其孩子最大直径比较的最大值。

```
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
int depth(TreeNode* root)
{
    if(!root) return 0;
    return 1+max(depth(root->left), depth(root->right));
}
public:
    int diameterOfBinaryTree(TreeNode* root) {
        if(!root || (!root->left && !root->right)) return 0;
        //cout<<depth(root->left)<< "+" <<depth(root->right)<<endl;
        int res=depth(root->left) +depth(root->right);
        res=max(res,diameterOfBinaryTree(root->right));
        res=max(res,diameterOfBinaryTree(root->left));
        return res;
        
    }
};
```

> 执行用时 :84 ms, 在所有 C++ 提交中击败了6.16%的用户\
> 内存消耗 :35.5 MB, 在所有 C++ 提交中击败了5.39%的用户

效率较低。另一种方法是找出某个根节点，该根节点的左右子树高度之和即为所求。\
使用深度优先搜索。

```
class Solution
{
public:
    int diameterOfBinaryTree(TreeNode* root)
    {
        int distance = 0;
        dfs(root, distance);
        return distance;
    }
    
    /**
     * distance 记录以root为根的子树的最大直径，返回树的高度
    */
    int dfs(TreeNode *root, int &distance)
    {
        if (root == nullptr)
            return 0;
        int left = dfs(root->left, distance);
        int right = dfs(root->right, distance);
        distance = max(left + right, distance);
        return max(left, right) + 1;
    }
};
```

> 执行用时 :20 ms, 在所有 C++ 提交中击败了67.87%的用户\
> 内存消耗 :19.6 MB, 在所有 C++ 提交中击败了89.71%的用户


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