求一个二叉树中两个节点的最近公共父节点

最新推荐文章于 2022-05-02 23:13:32 发布

原创最新推荐文章于 2022-05-02 23:13:32 发布 · 428 阅读

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java 同时被 2 个专栏收录

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本文介绍了一种使用递归方法寻找二叉树中两个指定节点的最近公共祖先的算法实现。通过构建特定的二叉树结构，并利用递归遍历左子树和右子树，最终找到同时包含这两个节点的最低层级的共同父节点。

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对树的考察在于使用递归方法（使用递归的方式）

package suanfa;

/**求一个二叉树中两个节点的最近公共父节点
*/
public class LowestCommonAncestor {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
//		TreeNode root = buildTree();
		TreeNode treeNode1 = new TreeNode(1, null, null);
		TreeNode treeNode2 = new TreeNode(2, null, null);
		TreeNode treeNode3 = new TreeNode(3, null, null);
		TreeNode treeNode4 = new TreeNode(4, null, null);
		TreeNode treeNode5 = new TreeNode(5, null, null);
		TreeNode treeNode6 = new TreeNode(6, null, null);
		treeNode1.left = treeNode2;
		treeNode1.right = treeNode3;
		treeNode2.left = treeNode4;
		treeNode2.right = treeNode5;
		treeNode4.right = treeNode6;
		TreeNode lowestCommonAncestor = getLowestCommonAncestor(treeNode1, treeNode5, treeNode6);
		System.out.println(lowestCommonAncestor.data);
		

		
			
			
	}
	
	/**
	 * 查找公共祖先
	 * @param root
	 * @param p
	 * @param q
	 * @return
	 */
	public static TreeNode  getLowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q) {
		if (root == null) return root;
		if (root == p || root == q ) return root;
		TreeNode leftN = getLowestCommonAncestor(root.left, p, q);
		TreeNode rightN = getLowestCommonAncestor(root.right, p, q);
		if (leftN != null && rightN !=null) return root;
		if (leftN == null) return rightN;
		return leftN;

		
		
	}
	
/*	public static TreeNode buildTree() {
		TreeNode treeNode1 = new TreeNode(1, null, null);
		TreeNode treeNode2 = new TreeNode(2, null, null);
		TreeNode treeNode3 = new TreeNode(3, null, null);
		TreeNode treeNode4 = new TreeNode(4, null, null);
		TreeNode treeNode5 = new TreeNode(5, null, null);
		TreeNode treeNode6 = new TreeNode(6, null, null);
		treeNode1.left = treeNode2;
		treeNode1.right = treeNode3;
		treeNode2.left = treeNode4;
		treeNode2.right = treeNode5;
		treeNode4.right = treeNode6;
		return treeNode1;
		
		
	}*/

}