687. 最长同值路径

English Version

题目描述

给定一个二叉树的 root ，返回 最长的路径的长度 ，这个路径中的 每个节点具有相同值 。 这条路径可以经过也可以不经过根节点。

两个节点之间的路径长度 由它们之间的边数表示。

 

示例 1:

输入：root = [5,4,5,1,1,5]
输出：2

示例 2:

输入：root = [1,4,5,4,4,5]
输出：2

 

提示:

• 树的节点数的范围是 [0, 104] 
• -1000 <= Node.val <= 1000
• 树的深度将不超过 1000 

解法

类似题目：543. 二叉树的直径

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def longestUnivaluePath(self, root: TreeNode) -> int:
        def dfs(root):
            if root is None:
                return 0
            left, right = dfs(root.left), dfs(root.right)
            left = left + 1 if root.left and root.left.val == root.val else 0
            right = right + 1 if root.right and root.right.val == root.val else 0
            nonlocal ans
            ans = max(ans, left + right)
            return max(left, right)

        ans = 0
        dfs(root)
        return ans

Java

/**
 * 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 {
    private int ans;

    public int longestUnivaluePath(TreeNode root) {
        ans = 0;
        dfs(root);
        return ans;
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = dfs(root.left);
        int right = dfs(root.right);
        left = root.left != null && root.left.val == root.val ? left + 1 : 0;
        right = root.right != null && root.right.val == root.val ? right + 1 : 0;
        ans = Math.max(ans, left + right);
        return Math.max(left, right);
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int ans;

    int longestUnivaluePath(TreeNode* root) {
        ans = 0;
        dfs(root);
        return ans;
    }

    int dfs(TreeNode* root) {
        if (!root) return 0;
        int left = dfs(root->left), right = dfs(root->right);
        left = root->left && root->left->val == root->val ? left + 1 : 0;
        right = root->right && root->right->val == root->val ? right + 1 : 0;
        ans = max(ans, left + right);
        return max(left, right);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func longestUnivaluePath(root *TreeNode) int {
	ans := 0
	var dfs func(root *TreeNode) int
	dfs = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		left, right := dfs(root.Left), dfs(root.Right)
		if root.Left != nil && root.Left.Val == root.Val {
			left++
		} else {
			left = 0
		}
		if root.Right != nil && root.Right.Val == root.Val {
			right++
		} else {
			right = 0
		}
		ans = max(ans, left+right)
		return max(left, right)
	}
	dfs(root)
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var longestUnivaluePath = function (root) {
    let ans = 0;
    let dfs = function (root) {
        if (!root) {
            return 0;
        }
        let left = dfs(root.left),
            right = dfs(root.right);
        left = root.left?.val == root.val ? left + 1 : 0;
        right = root.right?.val == root.val ? right + 1 : 0;
        ans = Math.max(ans, left + right);
        return Math.max(left, right);
    };
    dfs(root);
    return ans;
};

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