# 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)
