863. All Nodes Distance K in Binary Tree

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The key to solve this problem is to find the path from root to target, and put the length to target of every node from root to target to a map.

Then either using BFS or using DFS depends on you.

BFS:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<Integer> res = new ArrayList<>();
    public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        preOrder(root, target.val, map);
        bfs(root, k, map);
        return res;
    }
    
    private void bfs(TreeNode root, int k, Map<Integer, Integer> map){
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while(!queue.isEmpty()){
            int size = queue.size();
            for(int i=0;i<size;i++){
                TreeNode node = queue.poll();
                int len=0;
                if(map.containsKey(node.val)){
                    len = map.get(node.val);
                    if(len==k){
                        res.add(node.val);
                    }
                }
                if(node.left!=null){
                    if(!map.containsKey(node.left.val))
                        map.put(node.left.val, len+1);
                    queue.offer(node.left);
                }
                
                if(node.right!=null){
                    if(!map.containsKey(node.right.val))
                        map.put(node.right.val, len+1);
                    queue.offer(node.right);
                }
            }
        }
    }
    
    private int preOrder(TreeNode root, int target,  Map<Integer, Integer> map){
        if(root==null)
            return -1;
        if(root.val==target){
            map.put(root.val, 0);
            return 0;
        }
        int left = preOrder(root.left, target, map);
        int right = preOrder(root.right, target, map);
        if(left>=0){
            map.put(root.val, left+1);
            return left+1;
        }
        if(right>=0){
            map.put(root.val, right+1);
            return right+1;
        }
        return -1;
    }
}

DFS:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    List<Integer> res = new ArrayList<>();
    public List<Integer> distanceK(TreeNode root, TreeNode target, int k) {
        Map<Integer, Integer> map = new HashMap<>();
        preOrder(root, target.val, map);
        dfs(root, k, map);
        return res;
    }
    
    private void dfs(TreeNode root, int k, Map<Integer, Integer> map){
        if(map.get(root.val)==k){
            res.add(root.val);
        }
        if(root.left!=null){
            if(!map.containsKey(root.left.val)){
                map.put(root.left.val, map.get(root.val)+1);
            }
            dfs(root.left, k, map);
        }
        if(root.right!=null){
            if(!map.containsKey(root.right.val)){
                map.put(root.right.val, map.get(root.val)+1);
            }
            dfs(root.right, k, map);
        }
        
    }
    
    private int preOrder(TreeNode root, int target,  Map<Integer, Integer> map){
        if(root==null)
            return -1;
        if(root.val==target){
            map.put(root.val, 0);
            return 0;
        }
        int left = preOrder(root.left, target, map);
        int right = preOrder(root.right, target, map);
        if(left>=0){
            map.put(root.val, left+1);
            return left+1;
        }
        if(right>=0){
            map.put(root.val, right+1);
            return right+1;
        }
        return -1;
    }
}

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