# 236 Lowest Common Ancestor of a Binary Tree

## 236. Lowest Common Ancestor of a Binary Tree

## 1. Question

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](https://en.wikipedia.org/wiki/Lowest_common_ancestor): “The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow **a node to be a descendant of itself**).”

```
        _______3______
       /              \
    ___5__          ___1__
   /      \        /      \
   6      _2       0       8
         /  \
         7   4
```

For example, the lowest common ancestor (LCA) of nodes`5`and`1`is`3`. Another example is LCA of nodes`5`and`4`is`5`, since a node can be a descendant of itself according to the LCA definition.

## 2. Implementation

**(1) Recursion**

```java
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || root == p || root == q) {
            return root;
        }

        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        if (left != null && right != null) {
            return root;
        }
        return left != null ? left : right;
    }
}
```

**(2) Iteration**

```java
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        Map<TreeNode, TreeNode> parent = new HashMap<>();
        Stack<TreeNode> stack = new Stack<>();

        parent.put(root, null);
        stack.push(root);

        while (!parent.containsKey(p) || !parent.containsKey(q)) {
            TreeNode curNode = stack.pop();

            if (curNode.left != null) {
                parent.put(curNode.left, curNode);
                stack.push(curNode.left);
            }

            if (curNode.right != null) {
                parent.put(curNode.right, curNode);
                stack.push(curNode.right);
            }
        }

        Set<TreeNode> ancestor = new HashSet<>();
        while (p != null) {
            ancestor.add(p);
            p = parent.get(p);
        }

        while (!ancestor.contains(q)) {
            q = parent.get(q);
        }
        return q;
    }
}
```

## 3. Time & Space Complexity

**Recursion:** 时间复杂度O(n), 空间复杂度O(h)

**Iteration:** 时间复杂度O(n), 空间复杂度O(h)


---

# Agent Instructions: Querying This Documentation

If you need additional information that is not directly available in this page, you can query the documentation dynamically by asking a question.

Perform an HTTP GET request on the current page URL with the `ask` query parameter:

```
GET https://protegejj.gitbook.io/algorithm-practice/leetcode/tree/236-lowest-common-ancestor-of-a-binary-tree.md?ask=<question>
```

The question should be specific, self-contained, and written in natural language.
The response will contain a direct answer to the question and relevant excerpts and sources from the documentation.

Use this mechanism when the answer is not explicitly present in the current page, you need clarification or additional context, or you want to retrieve related documentation sections.