LeetCode 1740. Find Distance in a Binary Tree

原题链接在这里：https://leetcode.com/problems/find-distance-in-a-binary-tree/

题目：

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.

题解：

We could find the LCA.

And do a BFS from LCA to p and q to calculate the distance.

We could it in one pass as well. 

DFS returns current root distance to either p or q.

When root.val == p || q. One case is it is the first tiem we see p or q and didn't see the other, we return 0.

The other case is it is the second time we see p or q, which means we have seen both p and q. We find the LCA, it is either p or q. Return Math.max(l, r) + 1.

If both l >=0 && r >= 0, it means we find the LCA, its subtrees contains p and q. Return l + r + 2.

Time Complexity: O(n). n is the size of tree.

Space: O(logn).

AC Java:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode() {}
 8  *     TreeNode(int val) { this.val = val; }
 9  *     TreeNode(int val, TreeNode left, TreeNode right) {
10  *         this.val = val;
11  *         this.left = left;
12  *         this.right = right;
13  *     }
14  * }
15  */
16 class Solution {
17     int res = -1;
18     public int findDistance(TreeNode root, int p, int q) {
19         if(p == q){
20             return 0;
21         }
22         
23         dfs(root, p, q);
24         return res;
25     }
26     
27     private int dfs(TreeNode root, int p, int q){
28         if(root == null){
29             return -1;
30         }
31         
32         int l = dfs(root.left, p, q);
33         int r = dfs(root.right, p, q);
34         if(root.val == p || root.val == q){
35             if(l < 0 && r < 0){
36                 return 0;
37             }
38             
39             // we find the LCA.
40             res = Math.max(l, r) + 1;
41             return -1;
42         }
43         
44         if(l >= 0 && r >= 0){
45             // we find the LCA.
46             res = l + r + 2;
47             return -1;
48         }
49         
50         if(l >= 0){
51             return l + 1;
52         }
53         
54         if(r >= 0){
55             return r + 1;
56         }
57         
58         return -1;
59     }
60 }

AC Python:

 1 # Definition for a binary tree node.
 2 # class TreeNode:
 3 #     def __init__(self, val=0, left=None, right=None):
 4 #         self.val = val
 5 #         self.left = left
 6 #         self.right = right
 7 class Solution:
 8     def findDistance(self, root: Optional[TreeNode], p: int, q: int) -> int:
 9         def findLca(node, p, q):
10             if not node or node.val == p or node.val == q:
11                 return node
12             left = findLca(node.left, p, q)
13             right = findLca(node.right, p, q)
14             if left and right:
15                 return node
16             return left if left else right
17         
18         def findDis(node, target, dis):
19             if not node:
20                 return -1
21             if node.val == target:
22                 return dis
23             
24             left = findDis(node.left, target, dis + 1)
25             right = findDis(node.right, target, dis + 1)
26             if left != -1:
27                 return left
28             if right != -1:
29                 return right
30             return -1
31         
32         lca = findLca(root, p, q)
33         disP = findDis(lca, p, 0)
34         disQ = findDis(lca, q, 0)
35         return disP + disQ

类似Lowest Common Ancestor of a Binary Tree, Step-By-Step Directions From a Binary Tree Node to Another.

posted @ 2022-08-12 15:00  Dylan_Java_NYC  阅读(128)  评论(0)  编辑  收藏  举报