LeetCode Add One Row to Tree

LeetCode Add One Row to Tree

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   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:

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

在二叉树深度为d的那一层增加一行值全为v的节点,如果d=1的话,新增一个值为v的根节点,原来的树作为新根节点的左子树。

简单题,层次遍历到深度为d-1时,对d-1层的所有节点,都增加值为v的左右孩子,如果d-1层的节点原来有左右孩子,则原来的左右孩子接到新加入的v的节点上,作为左右孩子。

完整代码如下:

class Solution {
public:
	TreeNode* addOneRow(TreeNode* root, int v, int d) {
		if (d == 1) {
			TreeNode* newRoot = new TreeNode(v);
			newRoot->left = root;
			return newRoot;
		}

		queue<TreeNode*> tree;
		tree.push(root);
		int depth = 0;
		while (!tree.empty()) {
			++depth;
			int n = tree.size();
			if (depth == d - 1) {
				for (int i = 0; i < n; ++i) {
					TreeNode* p = tree.front();
					tree.pop();
					if (p == NULL)continue;

					TreeNode* l = new TreeNode(v);
					l->left = p->left;
					p->left = l;

					TreeNode* r = new TreeNode(v);
					r->right = p->right;
					p->right = r;
				}
				break;
			}
			else {
				for (int i = 0; i < n; ++i) {
					TreeNode* p = tree.front();
					tree.pop();
					if (p == NULL)continue;
					tree.push(p->left);
					tree.push(p->right);
				}
			}
		}
		return root;
	}
};

本代码提交AC,用时16MS。

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