[LeetCode] 236. Lowest Common Ancestor of a Binary Tree 二叉树的最近公共祖先
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]
_______3______
/ \
___5__ ___1__
/ \ / \
6 _2 0 8
/ \
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 of nodes5and1is3.
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 nodes5and4is5, 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.
与 235. Lowest Common Ancestor of a Binary Search Tree 类似，这道题是普通二叉树，不是二叉搜索树，所以不能利用二叉搜索树的性质，所以只能在二叉树中来搜索p和q，然后从路径中找到最近公共祖先。
解法：迭代
Java:
public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
if (root == null) return null;
if (root == p || root == q) return root;
TreeNode left = lowestCommonAncestor(root.left, p, q);
TreeNode right = lowestCommonAncestor(root.right, p, q);
if (left != null && right != null) return root;
else return (left != null) ? left : right;
}
}
Python:
class Solution:
# @param {TreeNode} root
# @param {TreeNode} p
# @param {TreeNode} q
# @return {TreeNode}
def lowestCommonAncestor(self, root, p, q):
if root in (None, p, q):
return root
left, right = [self.lowestCommonAncestor(child, p, q) \
for child in (root.left, root.right)]
# 1. If the current subtree contains both p and q,
# return their LCA.
# 2. If only one of them is in that subtree,
# return that one of them.
# 3. If neither of them is in that subtree,
# return the node of that subtree.
return root if left and right else left or right
C++：Recursion
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (!root || p == root || q == root) return root;
TreeNode *left = lowestCommonAncestor(root->left, p, q);
TreeNode *right = lowestCommonAncestor(root->right, p , q);
if (left && right) return root;
return left ? left : right;
}
};
C++:
class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (!root || root == p || root == q) {
return root;
}
TreeNode *left = lowestCommonAncestor(root->left, p, q);
TreeNode *right = lowestCommonAncestor(root->right, p, q);
// 1. If the current subtree contains both p and q,
// return their LCA.
// 2. If only one of them is in that subtree,
// return that one of them.
// 3. If neither of them is in that subtree,
// return the node of that subtree.
return left ? (right ? root : left) : right;
}
};
类似题目：
[LeetCode] 235. Lowest Common Ancestor of a Binary Search Tree 二叉搜索树的最近公共祖先
