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

    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:

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