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: โThe 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).โ
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
Time Complexity : O(N), where N is the number of nodes in the binary tree. In the worst case we might be visiting all the nodes of the binary tree.
Space Complexity : O(N). In the worst case space utilized by the stack, the parent pointer dictionary and the ancestor set, would be N each, since the height of a skewed binary tree could be N.
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */classSolution {publicTreeNodelowestCommonAncestor(TreeNode root,TreeNode p,TreeNode q) {Deque<TreeNode> stack =newArrayDeque<>();Map<TreeNode,TreeNode> parent =newHashMap<>();parent.put(root,null);stack.push(root);while (!parent.containsKey(p) ||!parent.containsKey(q)) {TreeNode node =stack.pop();if (node.left!=null) {parent.put(node.left, node);stack.push(node.left); }if (node.right!=null) {parent.put(node.right, node);stack.push(node.right); } }// Ancestros set() for node pSet<TreeNode> ancestors =newHashSet<>();// Process all ancestors for node p using parent pointerswhile (p !=null) {ancestrs.add(p); p =parent.get(p); }// the first ancestor of q which appears in p's ancestorwhile (!ancestors.contains(q)) { q =parent.get(q); }return q; }}