[LeetCode]993. 二叉树的堂兄弟节点(Java/c++ DFS)

Posted 2023-02-20 oyzg

Java/c++ DFS

链接：https://leetcode-cn.com/problems/cousins-in-binary-tree/solution/java-jie-jin-shuang-bai-by-oyzg-95kt/

解题思路
遍历树，找到x和y的深度和父节点即可

代码：
Java:

/**
 * 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 x_depth = 0;
	private int y_depth = 0;
	private TreeNode x_father = null;
	private TreeNode y_father = null;
	
	public boolean isCousins(TreeNode root, int x, int y) 
        dfs(root,x,y,0);
        return x_depth == y_depth && x_father != y_father;
    
	
	public void dfs(TreeNode root, int x, int y, int depth) 
		if(root == null) return ;
		if(root.left != null) 
			if(root.left.val == x) 
				x_depth = depth+1;
				x_father = root;
			
			if(root.left.val == y) 
				y_depth = depth+1;
				y_father = root;
			
			dfs(root.left,x,y,depth+1);
		
		if(root.right != null) 
			if(root.right.val == x) 
				x_depth = depth+1;
				x_father = root;
			
			if(root.right.val == y) 
				y_depth = depth+1;
				y_father = root;
			
			dfs(root.right,x,y,depth+1);
		
	

c++:

/**
 * 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 
private:
	int x_depth = 0;
	int y_depth = 0;
	TreeNode* x_father;
	TreeNode* y_father;
public:
    bool isCousins(TreeNode* root, int x, int y) 
        dfs(root,x,y,0);
        return x_depth == y_depth && x_father != y_father;
    
	void dfs(TreeNode* root, int x, int y, int depth) 
		if(root == nullptr) return ;
		if(root->left != nullptr) 
			if(root->left->val == x) 
				x_depth = depth+1;
				x_father = root;
			
			if(root->left->val == y) 
				y_depth = depth+1;
				y_father = root;
			
			dfs(root->left,x,y,depth+1);
		
		if(root->right != nullptr) 
			if(root->right->val == x) 
				x_depth = depth+1;
				x_father = root;
			
			if(root->right->val == y) 
				y_depth = depth+1;
				y_father = root;
			
			dfs(root->right,x,y,depth+1);
		
	
;