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Given the root of a binary tree, then value v and depth d, you need to add a row of nodes with value v at the given depth d. The root node is at depth 1.
The adding rule is: given a positive integer depth d, for each NOT null tree nodes N in depth d-1, create two tree nodes with value v as N's left subtree root and right subtree root. And N's original left subtree should be the left subtree of the new left subtree root, its original right subtree should be the right subtree of the new right subtree root. If depth d is 1 that means there is no depth d-1 at all, then create a tree node with value v as the new root of the whole original tree, and the original tree is the new root's left subtree.
Example 1:
Input: A binary tree as following: 4 / \ 2 6 / \ / 3 1 5 v = 1 d = 2 Output: 4 / \ 1 1 / \ 2 6 / \ / 3 1 5Example 2:
Input: A binary tree as following: 4 / 2 / \ 3 1 v = 1 d = 3 Output: 4 / 2 / \ 1 1 / \ 3 1Note:
The given d is in range [1, maximum depth of the given tree + 1].The given binary tree has at least one tree node.给定一个二叉树，根节点为第1层，深度为 1。在其第 d 层追加一行值为 v 的节点。
添加规则：给定一个深度值 d （正整数），针对深度为 d-1 层的每一非空节点 N，为 N 创建两个值为 v 的左子树和右子树。
将 N 原先的左子树，连接为新节点 v 的左子树；将 N 原先的右子树，连接为新节点 v 的右子树。
如果 d 的值为 1，深度 d - 1 不存在，则创建一个新的根节点 v，原先的整棵树将作为 v 的左子树。
示例 1:
输入: 二叉树如下所示: 4 / \ 2 6 / \ / 3 1 5 v = 1 d = 2 输出: 4 / \ 1 1 / \ 2 6 / \ / 3 1 5示例 2:
输入: 二叉树如下所示: 4 / 2 / \ 3 1 v = 1 d = 3 输出: 4 / 2 / \ 1 1 / \ 3 1注意:
输入的深度值 d 的范围是：[1，二叉树最大深度 + 1]。输入的二叉树至少有一个节点。44ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil 10 * self.right = nil 11 * } 12 * } 13 */ 14 class Solution { 15 func addOneRow(_ root: TreeNode?, _ v: Int, _ d: Int) -> TreeNode? { 16 if d == 1 { 17 let newRoot = TreeNode(v) 18 newRoot.left = root 19 return newRoot 20 } 21 helper(root, v, 1, d) 22 return root 23 } 24 25 private func helper(_ root: TreeNode?, _ v: Int, _ d: Int, _ targetD: Int) { 26 guard let root = root else { return } 27 if d == targetD - 1 { 28 let curLeft = root.left 29 let curRight = root.right 30 root.left = TreeNode(v) 31 root.right = TreeNode(v) 32 root.left?.left = curLeft 33 root.right?.right = curRight 34 return 35 } 36 helper(root.left, v, d + 1, targetD) 37 helper(root.right, v, d + 1, targetD) 38 } 39 }48ms
1 class Solution { 2 func addOneRow(_ root: TreeNode?, _ v: Int, _ d: Int) -> TreeNode? { 3 if root == nil { 4 return nil 5 } 6 7 var root = root 8 if d == 1 { 9 let newNode = TreeNode(v) 10 newNode.left = root 11 root = newNode 12 13 return root 14 } 15 16 var nodes = [(root, 1)] 17 while let (node, level) = nodes.popLast(), level < d { 18 if let left = node?.left { 19 nodes.insert((left, level + 1), at: 0) 20 } 21 if let right = node?.right { 22 nodes.insert((right, level + 1), at: 0) 23 } 24 25 if level == d - 1 { 26 let leftNode = TreeNode(v) 27 leftNode.left = node?.left 28 node?.left = leftNode 29 30 let rightNode = TreeNode(v) 31 rightNode.right = node?.right 32 node?.right = rightNode 33 } 34 35 if level > d { 36 break 37 } 38 } 39 40 return root 41 } 42 }60ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil 10 * self.right = nil 11 * } 12 * } 13 */ 14 15 class Solution { 16 func addOneRow(_ root: TreeNode?, _ v: Int, _ d: Int) -> TreeNode? { 17 if d == 1 { 18 var node = TreeNode(v) 19 node.left = root 20 return node 21 } 22 23 var stack = [TreeNode?]() 24 var tempArr = [TreeNode?]() 25 var currLevel = 1 26 stack.append(root) 27 while currLevel != d - 1 { 28 //print("currLevel: \(currLevel) ==============") 29 while !stack.isEmpty { 30 guard let node = stack.removeLast() else { 31 continue 32 } 33 //print("node: \(node.val)") 34 if let lNode = node.left { 35 tempArr.append(lNode) 36 } 37 if let rNode = node.right { 38 tempArr.append(rNode) 39 } 40 } 41 stack.append(contentsOf: tempArr) 42 tempArr.removeAll() 43 currLevel += 1 44 } 45 46 while !stack.isEmpty { 47 guard let node = stack.removeLast() else { 48 continue 49 } 50 //print("===== node: \(node.val)") 51 if let lNode = node.left { 52 var vNode = TreeNode(v) 53 vNode.left = lNode 54 node.left = vNode 55 } else { 56 var vNode = TreeNode(v) 57 node.left = vNode 58 } 59 if let rNode = node.right { 60 var vNode = TreeNode(v) 61 vNode.right = rNode 62 node.right = vNode 63 } else { 64 var vNode = TreeNode(v) 65 node.right = vNode 66 } 67 } 68 return root 69 } 70 }76ms
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * public var val: Int 5 * public var left: TreeNode? 6 * public var right: TreeNode? 7 * public init(_ val: Int) { 8 * self.val = val 9 * self.left = nil 10 * self.right = nil 11 * } 12 * } 13 */ 14 class Solution { 15 func addOneRow(_ root: TreeNode?, _ v: Int, _ d: Int) -> TreeNode? { 16 if d == 1 { 17 let node = TreeNode(v) 18 node.left = root 19 return node 20 } 21 22 23 if root == nil { 24 return nil 25 } 26 27 let ans = root 28 traverseAddOneRow(root!, v, d-1) 29 return ans 30 } 31 32 func traverseAddOneRow(_ root: TreeNode?, _ v: Int, _ d: Int) { 33 if root == nil { 34 return 35 } 36 37 38 if d < 1 { 39 return 40 } 41 42 if d == 1 { 43 44 let lastl = root!.left 45 let l = TreeNode(v) 46 root!.left = l 47 l.left = lastl 48 49 let lastr = root!.right 50 let r = TreeNode(v) 51 root!.right = r 52 r.right = lastr 53 54 return 55 } 56 traverseAddOneRow(root!.left, v, d-1) 57 traverseAddOneRow(root!.right, v, d-1) 58 } 59 60 }
转载于:https://www.cnblogs.com/strengthen/p/10472716.html
