LeetCode 236. Lowest Common Ancestor of a Binary Tree
Woodyiiiiiii opened this issue · comments
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 p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”
Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
Example 1:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.
Example 2:
Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
Note:
- All of the nodes' values will be unique.
- p and q are different and both values will exist in the binary tree.
实际上就是树形DP的问题。后序遍历,从左子树和右子树收集信息,到当前根节点处理。
处理信息分为几种情况:
- 左子树和右子树不为空,则该根节点就是最近公共祖先;
- 当前根节点为题目所给节点,左子树或右子树信息也返回的是所给节点,那么该根节点就是祖先;3. 左子树或右子树信息返回的是所给节点,而根节点不是,则返回所给节点;
- 当前根节点为题目所给节点,左子树或右子树信息返回空, 则返回所给节点;
- 其余情况为空;
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
return findLRC(root, p, q);
}
public TreeNode findLRC(TreeNode node, TreeNode p, TreeNode q) {
if (node == null) {
return null;
}
TreeNode left = findLRC(node.left, p, q);
TreeNode right = findLRC(node.right, p, q);
if ((left == p && right == q) ||
(left == q && right == p))
return node;
else if ((node == p || node == q)
&& (right != null || left != null))
return node;
else if (left != null || right != null)
return left == null ? right : left;
else if (node == p || node == q)
return node;
else
return null;
}
}
参考资料:
LeetCode原题