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algorithm notes: leetcode#543 diameter of binary tree

admin 11月 12, 2020 0

Problem

Solution

Initial thoughts

Python implementation

# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class :
    def diameterOfBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        def dfs(node):
            if node is None:
                return 0
            left = dfs(node.left)
            right = dfs(node.right)
            self.ans = max(self.ans, left+right+1)
            return max(left, right)+1
        self.ans = 1
        dfs(root)
        return self.ans-1

Java implementation

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class  {
    private int ans = 1;
    private int dfs(TreeNode node){
        if(node==null){ return 0; }
        int left = dfs(node.left);
        int right = dfs(node.right);
        int temp = left + right + 1;
        ans = temp > ans ? temp : ans;
        int depth = left > right ? left : right;
        return depth + 1;
    }
    public int diameterOfBinaryTree(TreeNode root) {
        dfs(root);
        return ans-1;
    }
}

Time complexity

O(N).

Space complexity

O(N).

Links

543. Diameter of Binary Tree
(中文版) 算法笔记: 力扣#543 二叉树的直径