Tree
Depth-First Search
Breadth-First Search
Hash Table
Binary Tree
Description
Given the root of a binary tree and two integers p
and q
, return the distance between the nodes of value p
and value q
in the tree .
The distance between two nodes is the number of edges on the path from one to the other.
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 0
Output: 3
Explanation: There are 3 edges between 5 and 0: 5-3-1-0.
Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 7
Output: 2
Explanation: There are 2 edges between 5 and 7: 5-2-7.
Example 3:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 5
Output: 0
Explanation: The distance between a node and itself is 0.
Constraints:
The number of nodes in the tree is in the range [1, 104 ]
.
0 <= Node.val <= 109
All Node.val
are unique .
p
and q
are values in the tree.
Solutions
Solution 1
Python3 Java C++ Go
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31 # 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 findDistance ( self , root : Optional [ TreeNode ], p : int , q : int ) -> int :
def lca ( root , p , q ):
if root is None or root . val in [ p , q ]:
return root
left = lca ( root . left , p , q )
right = lca ( root . right , p , q )
if left is None :
return right
if right is None :
return left
return root
def dfs ( root , v ):
if root is None :
return - 1
if root . val == v :
return 0
left , right = dfs ( root . left , v ), dfs ( root . right , v )
if left == right == - 1 :
return - 1
return 1 + max ( left , right )
g = lca ( root , p , q )
return dfs ( g , p ) + dfs ( g , q )
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51 /**
* 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 {
public int findDistance ( TreeNode root , int p , int q ) {
TreeNode g = lca ( root , p , q );
return dfs ( g , p ) + dfs ( g , q );
}
private int dfs ( TreeNode root , int v ) {
if ( root == null ) {
return - 1 ;
}
if ( root . val == v ) {
return 0 ;
}
int left = dfs ( root . left , v );
int right = dfs ( root . right , v );
if ( left == - 1 && right == - 1 ) {
return - 1 ;
}
return 1 + Math . max ( left , right );
}
private TreeNode lca ( TreeNode root , int p , int q ) {
if ( root == null || root . val == p || root . val == q ) {
return root ;
}
TreeNode left = lca ( root . left , p , q );
TreeNode right = lca ( root . right , p , q );
if ( left == null ) {
return right ;
}
if ( right == null ) {
return left ;
}
return root ;
}
}
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36 /**
* 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 findDistance ( TreeNode * root , int p , int q ) {
TreeNode * g = lca ( root , p , q );
return dfs ( g , p ) + dfs ( g , q );
}
TreeNode * lca ( TreeNode * root , int p , int q ) {
if ( ! root || root -> val == p || root -> val == q ) return root ;
TreeNode * left = lca ( root -> left , p , q );
TreeNode * right = lca ( root -> right , p , q );
if ( ! left ) return right ;
if ( ! right ) return left ;
return root ;
}
int dfs ( TreeNode * root , int v ) {
if ( ! root ) return -1 ;
if ( root -> val == v ) return 0 ;
int left = dfs ( root -> left , v );
int right = dfs ( root -> right , v );
if ( left == -1 && right == -1 ) return -1 ;
return 1 + max ( left , right );
}
};
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40 /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func findDistance ( root * TreeNode , p int , q int ) int {
var lca func ( root * TreeNode , p int , q int ) * TreeNode
lca = func ( root * TreeNode , p int , q int ) * TreeNode {
if root == nil || root . Val == p || root . Val == q {
return root
}
left , right := lca ( root . Left , p , q ), lca ( root . Right , p , q )
if left == nil {
return right
}
if right == nil {
return left
}
return root
}
var dfs func ( root * TreeNode , v int ) int
dfs = func ( root * TreeNode , v int ) int {
if root == nil {
return - 1
}
if root . Val == v {
return 0
}
left , right := dfs ( root . Left , v ), dfs ( root . Right , v )
if left == - 1 && right == - 1 {
return - 1
}
return 1 + max ( left , right )
}
g := lca ( root , p , q )
return dfs ( g , p ) + dfs ( g , q )
}
GitHub