993. Cousins in Binary Tree
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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
Note:
- The number of nodes in the tree will be between
2
and100
. - Each node has a unique integer value from
1
to100
.
class Solution { TreeNode xParent; TreeNode yParent; int xd = -1, yd = -1; public boolean isCousins(TreeNode root, int x, int y) { help(root, x, y, null, 0); return (xd == yd) && (xParent != yParent); } public void help(TreeNode root, int x, int y, TreeNode parent, int d){ if(root == null) return; if(root.val == x){ xParent = parent; xd = d; } if(root.val == y){ yParent = parent; yd = d; } help(root.left, x, y, root, d+1); help(root.right, x, y, root, d+1); } }
牢记前中后序遍历,depth的表达方法(下一层+1),以及parent的表示方法(从null开始,下一层parent就是这层的root)。
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