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

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Smallest Subtree with all the Deepest Nodes (M)

题目

Given the root of a binary tree, the depth of each node is the shortest distance to the root.

Return the smallest subtree such that it contains all the deepest nodes in the original tree.

A node is called the deepest if it has the largest depth possible among any node in the entire tree.

The subtree of a node is tree consisting of that node, plus the set of all descendants of that node.

Note: This question is the same as 1123: https://leetcode.com/problems/lowest-common-ancestor-of-deepest-leaves/

Example 1:

技术图片

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.

Example 2:

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

Example 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.

Constraints:

  • The number of nodes in the tree will be in the range [1, 500].
  • 0 <= Node.val <= 500
  • The values of the nodes in the tree are unique.

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。


代码实现

Java

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;
        }
    }
}

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

标签:get   inf   NPU   max   common   example   input   find   new   

原文地址:https://www.cnblogs.com/mapoos/p/14125334.html

版权声明:完美者 发表于 2020-12-17 12:55:47。
转载请注明:0865. Smallest Subtree with all the Deepest Nodes (M) | 完美导航

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