236. Lowest Common Ancestor of a Binary Tree

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.

M1: recursion

recursion + divide and conquer (左右两边分别递归）

base case: root为空，或者root为p或q

time: O(n), space: O(n)　　n: # nodes (worst case)

/**
 * 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) {
        if(root == null) return root;
        if(root == p || root == q) return root;
        
        TreeNode leftnode = lowestCommonAncestor(root.left, p, q);
        TreeNode rightnode = lowestCommonAncestor(root.right, p, q);
        
        if(leftnode != null && rightnode != null)
            return root;
        else if(rightnode == null)
            return leftnode;
        else if(leftnode == null)
            return rightnode;
        else
            return null;
    }
}

M2: iteration