543. Diameter of Binary Tree

Given a binary tree, you need to compute the length of the diameter of the tree. The diameter of a binary tree is the length of the longest path between any two nodes in a tree. This path may or may not pass through the root.

Example:
Given a binary tree 

          1
         / 
        2   3
       /      
      4   5    

Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3].

Note: The length of path between two nodes is represented by the number of edges between them.

//Time: O(n2), Space: O(h)
//这道题乍一眼看上去需要两个递归函数，对于每一个点要自己调用自己diameterOfBinaryTree，对于某一个点要调用dfs计算深度，但其实再仔细想不需要第一个递归，因为dfs在递归的同时已经可以计算出来周长了
    int diameter  = 0;
    
    public int diameterOfBinaryTree(TreeNode root) {
        if (root == null) {
            return 0;
        }

        dfs(root);
        return diameter;
    }
    
    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = dfs(root.left);
        int right = dfs(root.right);
        diameter = Math.max(diameter,left + right);//这里不加1， 因为周长算得是线段而不是点，比如只有一个node为1的点，周长为0.
        return 1 + Math.max(left,right);
    }