Leetcode 743. Closest Leaf in a Binary Tree

Posted 2020-10-19 Keep walking

Given a binary tree where every node has a unique value, and a target key k, find the value of the nearest leaf node to target k in the tree.

Here, nearest to a leaf means the least number of edges travelled on the binary tree to reach any leaf of the tree. Also, a node is called a leaf if it has no children.

In the following examples, the input tree is represented in flattened form row by row. The actual root tree given will be a TreeNode object.

Example 1:

Input:
root = [1, 3, 2], k = 1
Diagram of binary tree:
          1
         /         3   2

Output: 2 (or 3)

Explanation: Either 2 or 3 is the nearest leaf node to the target of 1.

 

Example 2:

Input:
root = [1], k = 1
Output: 1

Explanation: The nearest leaf node is the root node itself.

 

Example 3:

Input:
root = [1,2,3,4,null,null,null,5,null,6], k = 2
Diagram of binary tree:
             1
            /            2   3
          /
         4
        /
       5
      /
     6

Output: 3
Explanation: The leaf node with value 3 (and not the leaf node with value 6) is nearest to the node with value 2.

 

Note:

1. root represents a binary tree with at least 1 node and at most 1000 nodes.
2. Every node has a unique node.val in range [1, 1000].
3. There exists some node in the given binary tree for which node.val == k.

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     public int val;
 5  *     public TreeNode left;
 6  *     public TreeNode right;
 7  *     public TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int FindClosestLeaf(TreeNode root, int k) {
12         var leaves = new List<TreeNode>();
13         FindAllLeaves(root, leaves);
14         
15         var lcas = new List<TreeNode>();
16         foreach (var l in leaves)
17         {
18             lcas.Add(FindLCA(root, l.val, k));
19         }
20         
21         TreeNode res = null;
22         var min = Int32.MaxValue;
23         
24         for (int i = 0; i < lcas.Count; i++)
25         {
26             int cur = GetSteps(lcas[i], leaves[i].val) + GetSteps(lcas[i], k);
27             if (cur < min)
28             {
29                 res = leaves[i];
30                 min = cur;
31             }
32         }
33         
34         return res.val;
35     }
36     
37     private void FindAllLeaves(TreeNode node, IList<TreeNode> leaves)
38     {
39         if (node == null) return;
40         if (node.left == null && node.right == null)
41         {
42             leaves.Add(node);
43             return;
44         }
45         
46         FindAllLeaves(node.left, leaves);
47         FindAllLeaves(node.right, leaves);
48     }
49     
50     private TreeNode FindLCA(TreeNode node, int a, int b)
51     {
52         if (node == null) return null;
53         if (node.val == a || node.val == b) return node;
54         
55         var left = FindLCA(node.left, a, b);        
56         var right = FindLCA(node.right, a, b);
57         
58         if (left != null && right != null)
59         {
60             return node;
61         }
62         else if (left != null)
63         {
64             return left;
65         }
66         else
67         {
68             return right;
69         }
70     }
71     
72     private int GetSteps(TreeNode parent, int k)
73     {
74         if (parent == null) return -1;
75         if (parent.val == k) return 0;
76         
77         var left = GetSteps(parent.left, k);
78         if (left != -1)
79         {
80             return left + 1;
81         }
82         
83         var right = GetSteps(parent.right, k);
84         return right == -1 ? -1 : right + 1;
85     }
86 }