LeetCode 543. Diameter of Binary Tree

題目：

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.

---

又是一道不會做的easy，菜菜。

思路大概和max depth差不多，一邊求max depth，一邊計算以當前這個root為root的max diameter，然後和全域性的max做比較。複雜度都是O(n)。

Runtime: 0 ms, faster than 100.00% of Java online submissions for Diameter of Binary Tree.

Memory Usage: 38.8 MB, less than 14.95% of Java online submissions for Diameter of Binary Tree.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int numOfNodes = 0;
    public int diameterOfBinaryTree(TreeNode root) {
        if (root == null) {
            return 0;
        }
        depth(root);
        return numOfNodes - 1;
    }
    
    private int depth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = depth(root.left);
        int right = depth(root.right);
        numOfNodes = Math.max(numOfNodes, left + right + 1);
        return Math.max(left, right) + 1;
    }
}

另外一種做法和我的想法挺像的，就是root的max dia是左子樹的depth+右子樹的depth，然後遞迴求左右子節點的max dia，和root做比較。但是因為有重複計算所以效率挺低的。

Runtime: 9 ms, faster than 13.15% of Java online submissions for Diameter of Binary Tree.

Memory Usage: 38.7 MB, less than 14.95% of Java online submissions for Diameter of Binary Tree.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int diameterOfBinaryTree(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int rootDiameter = depth(root.left) + depth(root.right);
        int left = diameterOfBinaryTree(root.left);
        int right = diameterOfBinaryTree(root.right);
        return Math.max(rootDiameter, Math.max(left, right));
    }
    
    private int depth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = depth(root.left);
        int right = depth(root.right);
        return Math.max(left, right) + 1;
    }
}