LeetCode（算法）- 236. 二叉树的最近公共祖先

题目链接：​​点击打开链接​​

题目大意：略

解题思路：略

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AC 代码

• Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

// 解决方案(1)
class Solution {

    Queue<TreeNode> pQue = new LinkedList<>(), qQue = new LinkedList<>();

    Map<TreeNode, Integer> pMap = new HashMap<>(), qMap = new HashMap<>();

    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        dfs(root, p.val, pQue, pMap, 0);
        dfs(root, q.val, qQue, qMap, 0);
        TreeNode res = common();
        return res;
    }

    TreeNode common() {
        TreeNode res;
        while (!pQue.isEmpty() && !qQue.isEmpty()) {
            TreeNode pe = pQue.peek();
            TreeNode qe = qQue.peek();
            if (pe.val == qe.val) {
                res = pQue.peek();
                pQue.poll();
                qQue.poll();
                return res;
            }

            if (pMap.get(pe) > qMap.get(qe)) {
                pQue.poll();
            } else if (pMap.get(pe) < qMap.get(qe)) {
                qQue.poll();
            } else {
                pQue.poll();
                qQue.poll();
            }
        }
        return null;
    }

    boolean dfs(TreeNode node, int val, Queue<TreeNode> queue, Map<TreeNode, Integer> map, int l) {
        if (node == null) {
            return false;
        }

        if (node.val == val) {
            queue.offer(node);
            map.put(node, l);
            return true;
        }

        boolean left = dfs(node.left, val, queue, map, l + 1);
        if (left) {
            queue.offer(node);
            map.put(node, l);
            return true;
        }
        boolean right = dfs(node.right, val, queue, map, l + 1);
        if (right) {
            queue.offer(node);
            map.put(node, l);
            return true;
        }

        return false;
    }
}

// 解决方案(2)
class Solution {
    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);
        if(left == null) return right;
        if(right == null) return left;
        return root;
    }
}

• C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root == nullptr || root == p || root == q) return root;
        TreeNode *left = lowestCommonAncestor(root->left, p, q);
        TreeNode *right = lowestCommonAncestor(root->right, p, q);
        if(left == nullptr) return right;
        if(right == nullptr) return left;
        return root;
    }
};