Question

623. 在二叉树中增加一行

English Version

题目描述

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 <= 105
• 1 <= depth <= the depth of tree + 1

解法

方法一：DFS

# 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;
}

方法二：BFS

# 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;
}