[LeetCode 5.7] Cousins in Binary Tree

最新推荐文章于 2021-05-24 16:23:03 发布

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本文介绍了一种算法，用于判断二叉树中两个具有相同深度但不同父节点的节点是否为堂兄弟节点。通过查找指定值节点的父节点及其深度，可以有效地确定两节点是否满足堂兄弟条件。

Cousins in Binary Tree

In a binary tree, the root node is at depth 0, and children of each depth k node are at depth k+1.

Two nodes of a binary tree are cousins if they have the same depth, but have different parents.

We are given the root of a binary tree with unique values, and the values x and y of two different nodes in the tree.

Return true if and only if the nodes corresponding to the values x and y are cousins.

Example 1:

Input: root = [1,2,3,4], x = 4, y = 3
Output: false

Example 2:

Input: root = [1,2,3,null,4,null,5], x = 5, y = 4
Output: true

Example 3:

Input: root = [1,2,3,null,4], x = 2, y = 3
Output: false

Constraints:

The number of nodes in the tree will be between 2 and 100.
Each node has a unique integer value from 1 to 100.

Python3 Solution:

class Solution:
    def findval(self, root: TreeNode, val: int):
        if root.val == val:
            return None, 0
        depth = 1
        sta, nextsta = [root], []
        while sta:
            temp = sta.pop()
            if temp.left:
                if temp.left.val == val:
                    return temp, depth
                else:
                    nextsta.append(temp.left)
            if temp.right:
                if temp.right.val == val:
                    return temp, depth
                else:
                    nextsta.append(temp.right)
            if not sta:
                sta = nextsta[:]
                nextsta = []
                depth += 1
        return None, 0

    def isCousins(self, root: TreeNode, x: int, y: int) -> bool:
        node1, depth1 = self.findval(root, x)   # 找到x的父节点和深度
        node2, depth2 = self.findval(root, y)   # 找到y的父节点和深度
        if depth1 == depth2 and node1 != node2:
            return True
        else:
            return False