543 Diameter of Binary Tree

543. Diameter of Binary Tree​
1. Question
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
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          1
2
         / \
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        2   3
4
       / \     
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      4   5
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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.
2. Implementation
(1) DFS Recursion
思路: 这题其实和124找max path sum一样，基本的步骤都是先用postorder traversal 找出当前node的left path的信息，和right path的信息，然后用一个变量找出当前的信息加上左右两边信息的总和 与 当前已知的最大信息之间的较大者。返回的值则是当前的信息(节点的个数或者节点的值) 加上左右两边信息较大的一方
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/**
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 * Definition for a binary tree node.
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 * public class TreeNode {
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 *     int val;
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 *     TreeNode left;
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 *     TreeNode right;
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 *     TreeNode(int x) { val = x; }
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 * }
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 */
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class Solution {
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    public int diameterOfBinaryTree(TreeNode root) {
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        int[] diameter = new int[1];
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​
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        getDiameterOfTree(root, diameter);
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        return diameter[0];
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    }
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​
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    public int getDiameterOfTree(TreeNode node, int[] diameter) {
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        if (node == null) {
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            return 0;
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        }
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​
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        int leftDiameter = getDiameterOfTree(node.left, diameter);
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        int rightDiameter = getDiameterOfTree(node.right, diameter);
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        diameter[0] = Math.max(diameter[0], leftDiameter + rightDiameter);
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        return 1 + Math.max(leftDiameter, rightDiameter);
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    }
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}
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3. Time & Space Complexity
DFS Recursion: 时间复杂度O(n), n为树node的个数, 空间复杂度O(n)

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Contents

543. Diameter of Binary Tree

1. Question

2. Implementation

3. Time & Space Complexity

543. Diameter of Binary Tree​

1. Question

2. Implementation

3. Time & Space Complexity

543. Diameter of Binary Tree