leetcode Lowest Common Ancestor of a Binary Tree
1、binary search tree
Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.
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 binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]
Example 1:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8 Output: 6 Explanation: The LCA of nodes2
and8
is6
.
Example 2:
Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4 Output: 2 Explanation: The LCA of nodes2
and4
is2
, 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 BST.
当树为二叉搜索树时,当所找两节点值都大于(都小于)当前节点值时,就在右子树(左子树)中递归查找;
否则两个节点值分别大于、小于当前节点值或有一个等于当前节点值,当前节点即为公共祖先。
class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if (p->val > root->val && q->val > root->val) return lowestCommonAncestor(root->right, p, q); else if (p->val < root->val && q->val < root->val) return lowestCommonAncestor(root->left, p, q); else return root; } };
2、binary tree
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 nodes5
and1
is3.
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 nodes5
and4
is5
, 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.
当前节点为空或等于所找节点中的一个时,就返回当前节点;
否则分别在左子树和右子树中查找,若左右返回节点均非空时,由于树中无重复元素,所以所找两个节点分别在左子树和右子树中找到,当前节点就是公共祖先,返回当前节点。
class Solution { public: TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) { if (root == NULL || root == p || root == q) return root; auto l = lowestCommonAncestor(root->left, p, q); auto r = lowestCommonAncestor(root->right, p, q); if (l && r) return root; if (!l) return r; return l; } };