863. All Nodes Distance K in Binary Tree

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;
    }
}

标签：Distance,right,val,map,int,863,Tree,root,left
来源： https://www.cnblogs.com/feiflytech/p/16142917.html