0865. Smallest Subtree with all the Deepest Nodes (M)

题目

给定root二叉树的 ，每个节点的深度是到根的最短距离。

返回最小的子树，使其包含原始树中所有最深的节点。

如果一个节点在整个树中的任何节点中具有最大的可能深度，则称为最深节点。

节点的子树是由该节点加上该节点所有后代的集合组成的树。

注：本题同1123：https ://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/

示例 1：

Input: root = [3,5,1,6,2,0,8,null,null,7,4]
Output: [2,7,4]
Explanation: We return the node with value 2, colored in yellow in the diagram.
The nodes coloured in blue are the deepest nodes of the tree.
Notice that nodes 5, 3 and 2 contain the deepest nodes in the tree but node 2 is the smallest subtree among them, so we return it.

示例 2：

Input: root = [1]
Output: [1]
Explanation: The root is the deepest node in the tree.

示例 3：

Input: root = [0,1,3,null,2]
Output: [2]
Explanation: The deepest node in the tree is 2, the valid subtrees are the subtrees of nodes 2, 1 and 0 but the subtree of node 2 is the smallest.

约束：

• 树中的节点数将在范围内[1, 500]。
• 0 <= Node.val <= 500
• 树中节点的值是唯一的。

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题意

在二叉树中找到一个子树，使它包含所有深度最大的结点。

思路

先一次DFS找到所有最深的结点，然后用找公共祖先的方法递归处理：在当前结点的左子树和右子树中找包含最深结点的子树leftTree和rightTree，如果都存在，说明当前结点是一个公共祖先，返回该结点；如果只有leftTree存在，返回leftTree；如果只有rightTree存在，返回rightTree；如果都不存在，说明以当前结点为根结点的子树不在范围内，返回null。

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代码实现

爪哇

class Solution {
    private Map<TreeNode, Integer> depth = new HashMap<>();
    private int maxDepth = -1;

    public TreeNode subtreeWithAllDeepest(TreeNode root) {
        findDepth(root, 1);
        return findAncestor(root);
    }

    private void findDepth(TreeNode node, int level) {
        if (node == null) {
            return;
        }

        maxDepth = Math.max(maxDepth, level);
        depth.put(node, level);

        findDepth(node.left, level + 1);
        findDepth(node.right, level + 1);
    }

    private TreeNode findAncestor(TreeNode node) {
        if (node == null || depth.get(node) == maxDepth) {
            return node;
        }

        TreeNode leftAncestor = findAncestor(node.left);
        TreeNode rightAncestor = findAncestor(node.right);

        if (leftAncestor != null && rightAncestor != null) {
            return node;
        } else if (leftAncestor != null) {
            return leftAncestor;
        } else if (rightAncestor != null) {
            return rightAncestor;
        } else {
            return null;
        }
    }
}