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

## 236. Lowest Common Ancestor of a Binary Tree

## 题目地址

<https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/>

<http://www.lintcode.com/problem/lowest-common-ancestor/>

<http://www.jiuzhang.com/solutions/lowest-common-ancestor/>

## 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

```java
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.

```java
/**
 * 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) {
        ancestrs.add(p);
        p = parent.get(p);
      }

      // the first ancestor of q which appears in p's ancestor
      while (!ancestors.contains(q)) {
        q = parent.get(q);
      }

      return q;
    }
}
```


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