LeetCode 742. Closest Leaf in a Binary Tree

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

题目：

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.

题解：

First do DFS, find the TreeNode whose value is k, mark as kNode. At the same time, update HashMap with child->parent relationship.

Then do BFS from kNode, if any of left child, right child and parent is not null, put it into the queue. The first met leaf node is the nearest leaf node to kNode.

Time Complexity: O(n).

Space: O(n).

AC Java:

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 class Solution {
     TreeNode target;

     public int findClosestLeaf(TreeNode root, int k) {
         if(root == null){
             return -1;
         }

         HashMap<TreeNode, TreeNode> nodeToPar = new HashMap<>();
         dfs(root, null, nodeToPar, k);

         if(target == null){
             return -1;
         }

         LinkedList<TreeNode> que = new LinkedList<>();
         HashSet<Integer> visited = new HashSet<>();
         que.add(target);
         visited.add(k);

         while(!que.isEmpty()){
             TreeNode cur = que.poll();
             if(cur.left == null && cur.right == null){
                 return cur.val;
             }

             if(cur.left != null && !visited.contains(cur.left.val)){
                 que.add(cur.left);
                 visited.add(cur.left.val);
             }

             if(cur.right != null && !visited.contains(cur.right.val)){
                 que.add(cur.right);
                 visited.add(cur.right.val);
             }

             if(nodeToPar.containsKey(cur) && !visited.contains(nodeToPar.get(cur).val)){
                 que.add(nodeToPar.get(cur));
                 visited.add(nodeToPar.get(cur).val);
             }
         }

         return -1;
     }

     private void dfs(TreeNode root, TreeNode parent, HashMap<TreeNode, TreeNode> nodeToPar, int k){
         if(root == null){
             return;
         }

         if(parent != null){
             nodeToPar.put(root, parent);
         }

         if(root.val == k){
             target = root;
         }

         dfs(root.left, root, nodeToPar, k);
         dfs(root.right, root, nodeToPar, k);
     }
 }