# 236-lowest-common-ancestor-of-a-binary-tree


Try it on <a href='https://leetcode.com/problems/236-lowest-common-ancestor-of-a-binary-tree'>leetcode</a>

## Description
<div class="description">
<div><p>Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.</p>

<p>According to the <a href="https://en.wikipedia.org/wiki/Lowest_common_ancestor" target="_blank">definition of LCA on Wikipedia</a>: “The lowest common ancestor is defined between two nodes <code>p</code> and <code>q</code> as the lowest node in <code>T</code> that has both <code>p</code> and <code>q</code> as descendants (where we allow <b>a node to be a descendant of itself</b>).”</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;">
<pre><strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
<strong>Output:</strong> 3
<strong>Explanation:</strong> The LCA of nodes 5 and 1 is 3.
</pre>

<p><strong>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2018/12/14/binarytree.png" style="width: 200px; height: 190px;">
<pre><strong>Input:</strong> root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
<strong>Output:</strong> 5
<strong>Explanation:</strong> The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
</pre>

<p><strong>Example 3:</strong></p>

<pre><strong>Input:</strong> root = [1,2], p = 1, q = 2
<strong>Output:</strong> 1
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[2, 10<sup>5</sup>]</code>.</li>
	<li><code>-10<sup>9</sup> &lt;= Node.val &lt;= 10<sup>9</sup></code></li>
	<li>All <code>Node.val</code> are <strong>unique</strong>.</li>
	<li><code>p != q</code></li>
	<li><code>p</code> and <code>q</code> will exist in the tree.</li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class Solution:

    def __init__(self):
        # Variable to store LCA node.
        self.ans = None

    def lowestCommonAncestor(self, root, p, q):
        """
        :type root: TreeNode
        :type p: TreeNode
        :type q: TreeNode
        :rtype: TreeNode
        """
        def recurse_tree(current_node):

            # If reached the end of a branch, return False.
            if not current_node:
                return False

            # Left Recursion
            left = recurse_tree(current_node.left)

            # Right Recursion
            right = recurse_tree(current_node.right)

            # If the current node is one of p or q
            mid = current_node == p or current_node == q

            # If any two of the three flags left, right or mid become True.
            if mid + left + right >= 2:
                self.ans = current_node

            # Return True if either of the three bool values is True.
            return mid or left or right

        # Traverse the tree
        recurse_tree(root)
        return self.ans
```