[Leetcode] 623. Add One Row to Tree 解题报告

最新推荐文章于 2024-07-09 22:06:41 发布

魔豆Magicbean

最新推荐文章于 2024-07-09 22:06:41 发布

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分类专栏： IT公司面试习题文章标签：解题报告 Leetcode Tree

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本文介绍了一种在二叉树的特定深度插入一行节点的方法。通过递归方式，在指定深度d处添加值为v的新节点，同时确保原有节点结构正确调整。提供了详细的实现思路与C++代码示例。

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题目：

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 vas 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   5

Example 2:

Input: 
A binary tree as following:
      4
     /   
    2    
   / \   
  3   1    

v = 1

d = 3

Output: 
      4
     /   
    2
   / \    
  1   1
 /     \  
3       1

Note:

The given d is in range [1, maximum depth of the given tree + 1].
The given binary tree has at least one tree node.

思路：

思路还是递归，但这里特别需要注意的是递归的两个边界条件：1）d == 1，说明我们需要在当前层增加结点，并且将以当前结点为根的树都下降一层。2）当前结点为空，那么必须返回。值得注意的是我们需要首先判断条件1），因为在d == 1的时候，如果当前结点为空，我们也需要处理。如果前两个条件都没有满足，即d >= 2并且root ！= NULL，那么只需要递归调用，最后返回更新后的root即可。

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* addOneRow(TreeNode* root, int v, int d) {
        return addOneRow(root, v, d, true);
    }
private:
    TreeNode* addOneRow(TreeNode* node, int v, int d, bool is_left) {
        if (d == 1) {
            TreeNode *new_node = new TreeNode(v);
            if (is_left) {
                new_node->left = node;
            }
            else {
                new_node->right = node;
            }
            return new_node;
        }
        if (node == NULL) {     // d >= 2, so we return if node == NULL
            return NULL;
        }
        // node is not NULL and d >= 2, so we use recurrsion
        node->left = addOneRow(node->left, v, d - 1, true);
        node->right = addOneRow(node->right, v, d - 1, false);
        return node;
    }
};