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README.md

298. 二叉树最长连续序列

English Version

题目描述

给你一棵指定的二叉树的根节点 root ，请你计算其中 最长连续序列路径 的长度。

最长连续序列路径 是依次递增 1 的路径。该路径，可以是从某个初始节点到树中任意节点，通过「父 - 子」关系连接而产生的任意路径。且必须从父节点到子节点，反过来是不可以的。

 

示例 1：

输入：root = [1,null,3,2,4,null,null,null,5]
输出：3
解释：当中，最长连续序列是 3-4-5 ，所以返回结果为 3 。

示例 2：

输入：root = [2,null,3,2,null,1]
输出：2
解释：当中，最长连续序列是 2-3 。注意，不是 3-2-1，所以返回 2 。

 

提示：

  • 树中节点的数目在范围 [1, 3 * 104] 内
  • -3 * 104 <= Node.val <= 3 * 104

解法

DFS。

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 longestConsecutive(self, root: TreeNode) -> int:
        def dfs(root, p, t):
            nonlocal ans
            if root is None:
                return
            t = t + 1 if p is not None and p.val + 1 == root.val else 1
            ans = max(ans, t)
            dfs(root.left, root, t)
            dfs(root.right, root, t)

        ans = 1
        dfs(root, None, 1)
        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 longestConsecutive(TreeNode root) {
        ans = 1;
        dfs(root, null, 1);
        return ans;
    }

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

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 longestConsecutive(TreeNode* root) {
        ans = 1;
        dfs(root, nullptr, 1);
        return ans;
    }

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

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func longestConsecutive(root *TreeNode) int {
	ans := 1
	var dfs func(root, p *TreeNode, t int)
	dfs = func(root, p *TreeNode, t int) {
		if root == nil {
			return
		}
		if p != nil && p.Val+1 == root.Val {
			t++
			ans = max(ans, t)
		} else {
			t = 1
		}
		dfs(root.Left, root, t)
		dfs(root.Right, root, t)
	}
	dfs(root, nil, 1)
	return ans
}

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

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