LeetCode236 lowest Common Ancestor of a binary tree(二叉樹的最近公共祖先)
阿新 • • 發佈:2021-10-13
題目
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).
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. Example 3: Input: root = [1,2], p = 1, q = 2 Output: 1 Constraints: The number of nodes in the tree is in the range [2, 10⁵]. -10⁹ <= Node.val <= 10⁹ All Node.val are unique. p != q p and q will exist in the tree.
方法
深度優先遍歷法(DFS)
- 時間複雜度:O(n)
- 空間複雜度:O(n)
class Solution { private TreeNode result = null; public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) { dfs(root,p,q); return result; } private boolean dfs(TreeNode root, TreeNode p, TreeNode q){ if(root==null){ return false; } boolean left = dfs(root.left,p,q); boolean right = dfs(root.right,p,q); if(left&&right || ((left||right)&&(root.val==p.val||root.val==q.val))) { result = root; } return left || right || (root.val==p.val || root.val==q.val); } }
儲存父節點法
- 時間複雜度:O(n)
- 空間複雜度:O(n)