863. All Nodes Distance K in Binary Tree

最新推荐文章于 2020-12-23 13:14:13 发布

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最新推荐文章于 2020-12-23 13:14:13 发布

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分类专栏： leetcode

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本文介绍了一种算法，该算法接收一个二叉树（根节点为root）、目标节点target及整数K作为输入，返回树中所有与目标节点距离为K的节点值。示例显示了当输入特定二叉树、目标节点5和K值2时，如何获得输出[7,4,1]。文章提供了详细的Python实现，并解释了递归遍历树的方法。

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We are given a binary tree (with root node root), a target node, and an integer value `K`.

Return a list of the values of all nodes that have a distance K from the target node. The answer can be returned in any order.

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], target = 5, K = 2
Output: [7,4,1]
Explanation: 
The nodes that are a distance 2 from the target node (with value 5)
have values 7, 4, and 1.

Note that the inputs "root" and "target" are actually TreeNodes.
The descriptions of the inputs above are just serializations of these objects.

Note:

The given tree is non-empty.
Each node in the tree has unique values 0 <= node.val <= 500.
The target node is a node in the tree.
0 <= K <= 1000.

brute force，求出root到所有node的path，然后求距离

# Definition for a binary tree node.
class TreeNode:
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution:
    def distanceK(self, root, t, K):
        """
        :type root: TreeNode
        :type target: TreeNode
        :type K: int
        :rtype: List[TreeNode]
        """
        p, d = [], {}
        def dfs(r):
            if not r: return
            p.append(r)
            d[r] = list(p)
            dfs(r.left)
            dfs(r.right)
            p.pop()
        dfs(root)
        
        res=[]
        for k in d:
            i=0
            while i<len(d[k]) and i<len(d[t]) and d[k][i]==d[t][i]:
                i+=1
            if len(d[k])-i + len(d[t])-i == K:
                res.append(k.val)
        
        return res