Tree
Depth-First Search
Binary Tree
Description
Given the root
of a binary tree, return the length of the longest consecutive sequence path .
A consecutive sequence path is a path where the values increase by one along the path.
Note that the path can start at any node in the tree, and you cannot go from a node to its parent in the path.
Example 1:
Input: root = [1,null,3,2,4,null,null,null,5]
Output: 3
Explanation: Longest consecutive sequence path is 3-4-5, so return 3.
Example 2:
Input: root = [2,null,3,2,null,1]
Output: 2
Explanation: Longest consecutive sequence path is 2-3, not 3-2-1, so return 2.
Constraints:
The number of nodes in the tree is in the range [1, 3 * 104 ]
.
-3 * 104 <= Node.val <= 3 * 104
Solutions
Solution 1
Python3 Java C++ Go TypeScript
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25 # 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 : Optional [ TreeNode ]) -> int :
def dfs ( root : Optional [ TreeNode ]) -> int :
if root is None :
return 0
l = dfs ( root . left ) + 1
r = dfs ( root . right ) + 1
if root . left and root . left . val - root . val != 1 :
l = 1
if root . right and root . right . val - root . val != 1 :
r = 1
t = max ( l , r )
nonlocal ans
ans = max ( ans , t )
return t
ans = 0
dfs ( root )
return ans
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40 /**
* 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 ) {
dfs ( root );
return ans ;
}
private int dfs ( TreeNode root ) {
if ( root == null ) {
return 0 ;
}
int l = dfs ( root . left ) + 1 ;
int r = dfs ( root . right ) + 1 ;
if ( root . left != null && root . left . val - root . val != 1 ) {
l = 1 ;
}
if ( root . right != null && root . right . val - root . val != 1 ) {
r = 1 ;
}
int t = Math . max ( l , r );
ans = Math . max ( ans , t );
return t ;
}
}
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35 /**
* 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 longestConsecutive ( TreeNode * root ) {
int ans = 0 ;
function < int ( TreeNode * ) > dfs = [ & ]( TreeNode * root ) {
if ( ! root ) {
return 0 ;
}
int l = dfs ( root -> left ) + 1 ;
int r = dfs ( root -> right ) + 1 ;
if ( root -> left && root -> left -> val - root -> val != 1 ) {
l = 1 ;
}
if ( root -> right && root -> right -> val - root -> val != 1 ) {
r = 1 ;
}
int t = max ( l , r );
ans = max ( ans , t );
return t ;
};
dfs ( root );
return ans ;
}
};
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29 /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func longestConsecutive ( root * TreeNode ) ( ans int ) {
var dfs func ( * TreeNode ) int
dfs = func ( root * TreeNode ) int {
if root == nil {
return 0
}
l := dfs ( root . Left ) + 1
r := dfs ( root . Right ) + 1
if root . Left != nil && root . Left . Val - root . Val != 1 {
l = 1
}
if root . Right != nil && root . Right . Val - root . Val != 1 {
r = 1
}
t := max ( l , r )
ans = max ( ans , t )
return t
}
dfs ( root )
return
}
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35 /**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function longestConsecutive ( root : TreeNode | null ) : number {
let ans = 0 ;
const dfs = ( root : TreeNode | null ) : number => {
if ( root === null ) {
return 0 ;
}
let l = dfs ( root . left ) + 1 ;
let r = dfs ( root . right ) + 1 ;
if ( root . left && root . left . val - root . val !== 1 ) {
l = 1 ;
}
if ( root . right && root . right . val - root . val !== 1 ) {
r = 1 ;
}
const t = Math . max ( l , r );
ans = Math . max ( ans , t );
return t ;
};
dfs ( root );
return ans ;
}
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