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# 236.Lowest-Common-Ancestor-of-a-Binary-Tree

## 题目描述

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

## 代码

### Approach 1: Divide & Conquer

public class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode node1, TreeNode node2) {
if (root == null || root == node1 || root == node2) return root;
// Divide
TreeNode left = lowestCommonAncestor(root.left, node1, node2);
TreeNode right = lowestCommonAncestor(root.right, node1, node2);
// Conquer
if (left != null && right != null) {
return root;
}
if (left != null) {
return left;
}
if (right != null) {
return right;
}
return null;
}
}

### Approach #2 Itervative using parent points

Complexity Analysis
• 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; }
* }
*/
class Solution {
public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
Deque<TreeNode> stack = new ArrayDeque<>();
Map<TreeNode, TreeNode> parent = new HashMap<>();
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 p
Set<TreeNode> ancestors = new HashSet<>();
// Process all ancestors for node p using parent pointers
while (p != null) {