236. 二叉樹的最近公共祖先（樹）

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode ans = null;
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        findSons(root,p,q);
        return ans;
    }

    public boolean findSons(TreeNode root, TreeNode p, TreeNode q){
        if(root == null) return false;

        int left = findSons(root.left,p,q)?1:0;

        int right = findSons(root.right,p,q)?1:0;

        int mid = (root == p||root == q)?1:0;

        if(mid + left +right >1) ans = root;

        return(mid + left + right >0);
    }

}

方法二
將p，q的所有父節點都存入hashmap中，然後將p的所有父結點放入set；遍歷q的所有父結點，第一個匹配set的即爲公共祖先

class Solution {

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {

        // Stack for tree traversal
        Deque<TreeNode> stack = new ArrayDeque<>();

        // HashMap for parent pointers
        Map<TreeNode, TreeNode> parent = new HashMap<>();

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

        // Iterate until we find both the nodes p and q
        while (!parent.containsKey(p) || !parent.containsKey(q)) {

            TreeNode node = stack.pop();

            // While traversing the tree, keep saving the parent pointers.
            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);
            }
        }

        // Ancestors set() for node p.
        Set<TreeNode> ancestors = new HashSet<>();

        // Process all ancestors for node p using parent pointers.
        while (p != null) {
            ancestors.add(p);
            p = parent.get(p);
        }

        // The first ancestor of q which appears in
        // p's ancestor set() is their lowest common ancestor.
        while (!ancestors.contains(q))
            q = parent.get(q);
        return q;
    }

}