236. Lowest Common Ancestor of a Binary Tree

最新推荐文章于 2024-02-06 06:38:28 发布

mingyangdai

最新推荐文章于 2024-02-06 06:38:28 发布

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本文介绍了一种寻找二叉树中两个指定节点的最低公共祖先(LCA)的算法。通过两种方法实现：一种利用栈和哈希映射进行迭代；另一种采用递归方式，更为高效。详细解析了每种方法的工作原理。

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself).”

_______3______

/ \

___5__ ___1__

/ \ / \

6 _2 0 8

/ \

7 4

For example, the lowest common ancestor (LCA) of nodes 5 and 1 is 3. Another example is LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

找两棵树的最低父节点

自己做出来了一个递归的解法但是stackoverflow了 testcase的树深度特别大下面的解法会占用大量内存但是可以通过testcase

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    Map<TreeNode, TreeNode> parent = new HashMap<>();
    Deque<TreeNode> stack = new ArrayDeque<>();
    parent.put(root, null);
    stack.push(root);

    while (!parent.containsKey(p) || !parent.containsKey(q)) {
        TreeNode node = stack.pop();
        if (node.left != null) {
            parent.put(node.left, node);
            stack.push(node.left);
        }
        if (node.right != null) {
            parent.put(node.right, node);
            stack.push(node.right);
        }
    }
    Set<TreeNode> ancestors = new HashSet<>();
    while (p != null) {
        ancestors.add(p);
        p = parent.get(p);
    }
    while (!ancestors.contains(q))
        q = parent.get(q);
    return q;
}

遍历整棵树把对应关系存入parent key是节点 value是父节点

然后获取p的所有父节点对q 从下到上逐层获取父节点找到第一个与p的公共父节点也就是最低的公共父节点

下面是递归的解法

public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
    if(root == null || root == p || root == q)  return root;//1.
    TreeNode left = lowestCommonAncestor(root.left, p, q);
    TreeNode right = lowestCommonAncestor(root.right, p, q);
    if(left != null && right != null)   return root;//2.
    return left != null ? left : right;//3.
}

效率比较高但是之前觉得不是很好理解

四个月之后再看感觉也不难理解

1.如果p是根节点那么直接返回因为q肯定是p的子节点

2.如果在当前节点下找到了p和q 那当前节点就是最低父节点返回即可

3.如果在当前节点下没找到q 那说明q是p的子节点返回p即可