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index.js
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/*
* @lc app=leetcode id=687 lang=javascript
*
* [687] Longest Univalue Path
*
* https://leetcode.com/problems/longest-univalue-path/description/
*
* algorithms
* Easy (33.21%)
* Total Accepted: 51.4K
* Total Submissions: 154.1K
* Testcase Example: '[5,4,5,1,1,5]'
*
* Given a binary tree, find the length of the longest path where each node in
* the path has the same value. This path may or may not pass through the
* root.
*
* Note: The length of path between two nodes is represented by the number of
* edges between them.
*
*
* Example 1:
*
*
*
*
* Input:
*
* 5
* / \
* 4 5
* / \ \
* 1 1 5
*
*
*
*
* Output:
*
* 2
*
*
*
*
* Example 2:
*
*
*
*
* Input:
*
* 1
* / \
* 4 5
* / \ \
* 4 4 5
*
*
*
*
* Output:
*
* 2
*
*
*
* Note:
* The given binary tree has not more than 10000 nodes. The height of the tree
* is not more than 1000.
*
*/
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
const { make_tree } = require('../utils/');
const root = make_tree([5,4,5,1,1,null, 5], [0])[0];
/**
* 思路:
* 1.递归的找经过每个节点的最大路径
* 2.返回当前节点的最大路径,如果left的值和root一样则leftDepth + 1否则为零,如果right的值和root一样则rightDepth + 1否则为零,
* 3.返回的时候只能返回左或者右的最大值
*
* @param {TreeNode} root
* @return {number}
*/
var longestUnivaluePath = function(root) {
let maxLength = 0;
dfs(root);
return maxLength;
function dfs(node) {
if (!node) {
return 0;
}
let leftMaxDepth = dfs(node.left);
let rightMaxDepth = dfs(node.right);
if (
node.left &&
node.left.val === node.val
) {
leftMaxDepth++;
} else {
leftMaxDepth = 0;
}
if (
node.right &&
node.right.val === node.val
) {
rightMaxDepth++;
} else {
rightMaxDepth = 0;
}
const maxSubDepth = Math.max(leftMaxDepth, rightMaxDepth);
maxLength = Math.max(maxLength, leftMaxDepth + rightMaxDepth);
return maxSubDepth;
}
};
longestUnivaluePath(root);
module.exports = {
id:'687',
title:'Longest Univalue Path',
url:'https://leetcode.com/problems/longest-univalue-path/description/',
difficulty:'Easy',
}