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:
rootrepresents a binary tree with at least1node and at most1000nodes.- Every node has a unique
node.valin range[1, 1000]. - 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 }
