236. Lowest Common Ancestor of a Binary Tree

原题链接：https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
代码实现如下：

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

/**
 * Created by clearbug on 2018/2/26.
 */
public class Solution {

    public static void main(String[] args) {
        Solution s = new Solution();
    }

    // 讨论区一位高手的答案，不得不说：高手就是高手啊！
    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);
        return left == null ? right : right == null ? left : root;
    }


    /**
     * 方法一：这是我自己想的方法，何其复杂啊！！！提交之后运行结果：StackOverFlowException
     *
     * @param root
     * @param p
     * @param q
     * @return
     */
    public TreeNode lowestCommonAncestor1(TreeNode root, TreeNode p, TreeNode q) {
        if (root == null || p == null || q == null) {
            return null;
        }

        Stack<TreeNode> pStack = new Stack<>();
        Stack<TreeNode> qStack = new Stack<>();

        if (!getNodePath(root, p, pStack)) {
            return null;
        }
        if (!getNodePath(root, q, qStack)) {
            return null;
        }

        Queue<TreeNode> pQueue = new LinkedList<>(pStack);
        Queue<TreeNode> qQueue = new LinkedList<>(qStack);

        TreeNode lca = null;
        while (!pQueue.isEmpty() && !qQueue.isEmpty()) {
            if (pQueue.peek() != qQueue.peek()) {
                break;
            }
            lca = pQueue.peek();
            pQueue.poll();
            qQueue.poll();
        }

        return lca;
    }

    private boolean getNodePath(TreeNode root, TreeNode target, Stack<TreeNode> stack) {
        if (root == target) {
            return true;
        }

        stack.push(root);
        if (root.left != null) {
            if (getNodePath(root.left, target, stack)) {
                return true;
            } else {
                stack.pop();
            }
        }
        if (root.right != null) {
            if (getNodePath(root.right, target, stack)) {
                return true;
            } else {
                stack.pop();
            }
        }

        return false;
    }
}