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;
}
}
}