| comments | difficulty | edit_url | tags | ||||
|---|---|---|---|---|---|---|---|
true |
中等 |
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给定一个二叉树的根 root 和两个整数 val 和 depth ，在给定的深度 depth 处添加一个值为 val 的节点行。
注意，根节点 root 位于深度 1 。
加法规则如下:
- 给定整数
depth，对于深度为depth - 1的每个非空树节点cur，创建两个值为val的树节点作为cur的左子树根和右子树根。 cur原来的左子树应该是新的左子树根的左子树。cur原来的右子树应该是新的右子树根的右子树。- 如果
depth == 1意味着depth - 1根本没有深度，那么创建一个树节点，值val作为整个原始树的新根，而原始树就是新根的左子树。
示例 1:
输入: root = [4,2,6,3,1,5], val = 1, depth = 2 输出: [4,1,1,2,null,null,6,3,1,5]
示例 2:
输入: root = [4,2,null,3,1], val = 1, depth = 3 输出: [4,2,null,1,1,3,null,null,1]
提示:
- 节点数在
[1, 104]范围内 - 树的深度在
[1, 104]范围内 -100 <= Node.val <= 100-105 <= val <= 1051 <= depth <= the depth of tree + 1
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def addOneRow(
self, root: Optional[TreeNode], val: int, depth: int
) -> Optional[TreeNode]:
def dfs(root, d):
if root is None:
return
if d == depth - 1:
root.left = TreeNode(val, root.left, None)
root.right = TreeNode(val, None, root.right)
return
dfs(root.left, d + 1)
dfs(root.right, d + 1)
if depth == 1:
return TreeNode(val, root)
dfs(root, 1)
return root/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
private int val;
private int depth;
public TreeNode addOneRow(TreeNode root, int val, int depth) {
if (depth == 1) {
return new TreeNode(val, root, null);
}
this.val = val;
this.depth = depth;
dfs(root, 1);
return root;
}
private void dfs(TreeNode root, int d) {
if (root == null) {
return;
}
if (d == depth - 1) {
TreeNode l = new TreeNode(val, root.left, null);
TreeNode r = new TreeNode(val, null, root.right);
root.left = l;
root.right = r;
return;
}
dfs(root.left, d + 1);
dfs(root.right, d + 1);
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
int val;
int depth;
TreeNode* addOneRow(TreeNode* root, int val, int depth) {
if (depth == 1) return new TreeNode(val, root, nullptr);
this->val = val;
this->depth = depth;
dfs(root, 1);
return root;
}
void dfs(TreeNode* root, int d) {
if (!root) return;
if (d == depth - 1) {
auto l = new TreeNode(val, root->left, nullptr);
auto r = new TreeNode(val, nullptr, root->right);
root->left = l;
root->right = r;
return;
}
dfs(root->left, d + 1);
dfs(root->right, d + 1);
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func addOneRow(root *TreeNode, val int, depth int) *TreeNode {
if depth == 1 {
return &TreeNode{Val: val, Left: root}
}
var dfs func(root *TreeNode, d int)
dfs = func(root *TreeNode, d int) {
if root == nil {
return
}
if d == depth-1 {
l, r := &TreeNode{Val: val, Left: root.Left}, &TreeNode{Val: val, Right: root.Right}
root.Left, root.Right = l, r
return
}
dfs(root.Left, d+1)
dfs(root.Right, d+1)
}
dfs(root, 1)
return root
}/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function addOneRow(root: TreeNode | null, val: number, depth: number): TreeNode | null {
function dfs(root, d) {
if (!root) {
return;
}
if (d == depth - 1) {
root.left = new TreeNode(val, root.left, null);
root.right = new TreeNode(val, null, root.right);
return;
}
dfs(root.left, d + 1);
dfs(root.right, d + 1);
}
if (depth == 1) {
return new TreeNode(val, root);
}
dfs(root, 1);
return root;
}# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def addOneRow(
self, root: Optional[TreeNode], val: int, depth: int
) -> Optional[TreeNode]:
if depth == 1:
return TreeNode(val, root)
q = deque([root])
i = 0
while q:
i += 1
for _ in range(len(q)):
node = q.popleft()
if node.left:
q.append(node.left)
if node.right:
q.append(node.right)
if i == depth - 1:
node.left = TreeNode(val, node.left, None)
node.right = TreeNode(val, None, node.right)
return root/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode addOneRow(TreeNode root, int val, int depth) {
if (depth == 1) {
return new TreeNode(val, root, null);
}
Deque<TreeNode> q = new ArrayDeque<>();
q.offer(root);
int i = 0;
while (!q.isEmpty()) {
++i;
for (int k = q.size(); k > 0; --k) {
TreeNode node = q.pollFirst();
if (node.left != null) {
q.offer(node.left);
}
if (node.right != null) {
q.offer(node.right);
}
if (i == depth - 1) {
node.left = new TreeNode(val, node.left, null);
node.right = new TreeNode(val, null, node.right);
}
}
}
return root;
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* addOneRow(TreeNode* root, int val, int depth) {
if (depth == 1) return new TreeNode(val, root, nullptr);
queue<TreeNode*> q{{root}};
int i = 0;
while (!q.empty()) {
++i;
for (int k = q.size(); k; --k) {
TreeNode* node = q.front();
q.pop();
if (node->left) q.push(node->left);
if (node->right) q.push(node->right);
if (i == depth - 1) {
node->left = new TreeNode(val, node->left, nullptr);
node->right = new TreeNode(val, nullptr, node->right);
}
}
}
return root;
}
};/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func addOneRow(root *TreeNode, val int, depth int) *TreeNode {
if depth == 1 {
return &TreeNode{val, root, nil}
}
q := []*TreeNode{root}
i := 0
for len(q) > 0 {
i++
for k := len(q); k > 0; k-- {
node := q[0]
q = q[1:]
if node.Left != nil {
q = append(q, node.Left)
}
if node.Right != nil {
q = append(q, node.Right)
}
if i == depth-1 {
node.Left = &TreeNode{val, node.Left, nil}
node.Right = &TreeNode{val, nil, node.Right}
}
}
}
return root
}/**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function addOneRow(root: TreeNode | null, val: number, depth: number): TreeNode | null {
if (depth === 1) {
return new TreeNode(val, root);
}
const queue = [root];
for (let i = 1; i < depth - 1; i++) {
const n = queue.length;
for (let j = 0; j < n; j++) {
const { left, right } = queue.shift();
left && queue.push(left);
right && queue.push(right);
}
}
for (const node of queue) {
node.left = new TreeNode(val, node.left);
node.right = new TreeNode(val, null, node.right);
}
return root;
}