[leetCode]剑指 Offer 68

解法

从根节点开始，判断p，q是否存在于当前节点的左子树和右子树之中：

• 如果存在则当前节点为pq的最近公共节点
• 如果p q 在当前节点的左子树则遍历左子树
• 如果p q 在当前节点的右子树则遍历右子树

class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null)
            return null;
        if (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;
        else if (left != null)
            return left;
        else if (right != null)
            return right;
        return null;
    }
}

解法二

使用前序遍历将p, q节点所在的路径保存下来，再寻找两条路径上最后的公共节点

/**
 * 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) {
       LinkedList<TreeNode> path1 = new LinkedList<>();
       getNodePath(root, p, path1);
       LinkedList<TreeNode> path2 = new LinkedList<>();
       getNodePath(root, q, path2);
       return getLastCommonNode(path1, path2);
    }

    private boolean getNodePath(TreeNode root, TreeNode node, LinkedList<TreeNode> path) {
        if (root == node) {
            path.offer(root);
            return true;
        }
        if (root == null)
            return false;
        path.offer(root);
        boolean left = getNodePath(root.left, node, path);
        if (left) return true;
        boolean right = getNodePath(root.right, node, path);
        if (right) return true;
        path.pollLast();
        return false;
    }

    private TreeNode getLastCommonNode
    (
        LinkedList<TreeNode> path1,
        LinkedList<TreeNode> path2
    ) {
        TreeNode last = null;
        while (!path1.isEmpty() && !path2.isEmpty()) {
            TreeNode node1 = path1.pollFirst();
            TreeNode node2 = path2.pollFirst();
            if (node1 == node2) 
                last = node1;
        }
        return last;
    }
}