543- binary tree diameter

Suppose we know that for the left son of the node after traversing downwardly the largest number of nodes L (i.e., the left son of the root of the subtree depth) thereof and a right son is traversed downward through the most nodes R (i.e. to the right son the depth of the root of the subtree), then to the node as the starting point of the path passes through a maximum number of nodes is the L + R + 1

We note node node as the starting point of the path passes through a maximum number of nodes is dnode, then the binary tree diameter is the maximum of all nodes dnode minus one.

The final algorithm process: We define a recursive function depth (node) function returns the node calculates dnode depth of the root of the subtree. Recursive call to the left and right son son depth L and RR are determined subtree root, the depth of the root node of the subtree that is, max (L, R) +1; dnode value of the node is L + R + 1  recursively search every node and set a global variable record ans  dnode most large value, and finally return ans-1 is the diameter of the tree.

Link: https: //leetcode-cn.com/problems/diameter-of-binary-tree/solution/er-cha-shu-de-zhi-jing-by-leetcode-solution/

 1 /**
 2  * Definition for a binary tree node.
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int length=0;
13     int diameterOfBinaryTree(TreeNode* root) {
14         depth(root);
15         return length;
16     }
17     int depth(TreeNode* root)
18     {
19         int L=depth(root->left);
20         int R=depth(root-right);
21         length=max(L+R,length);
22         return max(L,R)+1;
23     }
24 };

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