vue父元件向子元件傳值,不實時更新解決
阿新 • • 發佈:2020-12-26
劍指 Offer 28. 對稱的二叉樹
請實現一個函式,用來判斷一棵二叉樹是不是對稱的。如果一棵二叉樹和它的映象一樣,那麼它是對稱的。
例如,二叉樹 [1,2,2,3,4,4,3] 是對稱的。
1
/
2 2
/ \ /
3 4 4 3
但是下面這個 [1,2,2,null,3,null,3] 則不是映象對稱的:1
/
2 2
\
3 3示例 1:
輸入:root = [1,2,2,3,4,4,3]
輸出:true
示例 2:輸入:root = [1,2,2,null,3,null,3]
輸出:false限制:
0 <= 節點個數 <= 1000
/** * Definition for a binary tree node. * class TreeNode(var _value: Int) { * var value: Int = _value * var left: TreeNode = null * var right: TreeNode = null * } */ object Solution { def isSymmetric(root: TreeNode): Boolean = { if (root == null) {return true} return recur(root.left, root.right) } def recur(A: TreeNode, B: TreeNode): Boolean = { if (A == null && B == null) {return true} if (A == null || B == null || A.value != B.value) {return false} return recur(A.left, B.right) && recur(A.right, B.left) } }
/** * Definition for a binary tree node. * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ func isSymmetric(root *TreeNode) bool { if root == nil {return true} return recur(root.Left, root.Right) } func recur(left *TreeNode, right *TreeNode) bool { if left == nil && right == nil {return true} if left == nil || right == nil || left.Val != right.Val {return false} return recur(left.Left, right.Right) && recur(left.Right, right.Left) }