742. Closest Leaf in a Binary Tree

最新推荐文章于 2024-05-04 15:02:13 发布

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本文介绍了一种算法，用于在给定的二叉树中找到离指定目标节点最近的叶子节点。通过遍历树结构并使用队列进行搜索，该方法能够有效地找出距离目标节点最近的叶子节点值。

Given a binary tree where every node has a unique value, and a target key k, find the value of the nearest leaf node to target k in the tree.

Here, nearest to a leaf means the least number of edges travelled on the binary tree to reach any leaf of the tree. Also, a node is called a leaf if it has no children.

In the following examples, the input tree is represented in flattened form row by row. The actual root tree given will be a TreeNode object.

Example 1:

Input:
root = [1, 3, 2], k = 1
Diagram of binary tree:
          1
         / \
        3   2

Output: 2 (or 3)

Explanation: Either 2 or 3 is the nearest leaf node to the target of 1.

Example 2:

Input:
root = [1], k = 1
Output: 1

Explanation: The nearest leaf node is the root node itself.

Example 3:

Input:
root = [1,2,3,4,null,null,null,5,null,6], k = 2
Diagram of binary tree:
             1
            / \
           2   3
          /
         4
        /
       5
      /
     6

Output: 3
Explanation: The leaf node with value 3 (and not the leaf node with value 6) is nearest to the node with value 2.

Note:

root represents a binary tree with at least 1 node and at most 1000 nodes.
Every node has a unique node.val in range [1, 1000].
There exists some node in the given binary tree for which node.val == k.

class Solution {
	public:
		int findClosestLeaf(TreeNode* root, int k) {
			kv=k;
			for(int i=0;i<1010;i++)
			{
				f[i]=false;
				v[i]=false;
			}
			helper(root);
			queue<int> q;
			q.push(k);
			int res=0;
			while(q.size()>0)
			{
				int tk=q.front();
				if(f[tk] == true)
				  return tk;
				q.pop();
				if( v[tk] == false)
				{
					v[tk]=true;
					res++;
					int ns=n[tk].size();
					for(int i=0;i<ns;i++)
					{
						int tki=n[tk][i];
						if(f[tki] == true)
						  return tki;
						if(v[tki] == false)
						{
							q.push(tki);
						}
					}
				}
			}
			return -1;
		}
		void helper(TreeNode* t)
		{
			if(t == NULL)
			  return;
			if(t->left == NULL && t->right == NULL)
			{
				f[t->val]=true;
				return;
			}
			if(t->left !=NULL)
			{
				n[t->left->val].push_back(t->val);
				n[t->val].push_back(t->left->val);
				helper(t->left);
			}
			if(t->right !=NULL)
			{
				n[t->right->val].push_back(t->val);
				n[t->val].push_back(t->right->val);
				helper(t->right);
			}
		}
	private:
		vector<int> n[1010];
		bool f[1010];
		bool v[1010];
		int kv;
};