LeetCode（剑指 Offer）- 68 - I. 二叉搜索树的最近公共祖先

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 {
 
    Stack<TreeNode> pStack = new Stack<>(), qStack = new Stack<>();
 
    Map<TreeNode, Integer> pMap = new HashMap<>(), qMap = new HashMap<>();
 
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        dfs(root, p.val, pStack, pMap, 0);
        dfs(root, q.val, qStack, qMap, 0);
        TreeNode res = common();
        return res;
    }
 
    TreeNode common() {
        TreeNode res;
        while (!pStack.isEmpty() && !qStack.isEmpty()) {
            TreeNode pe = pStack.peek();
            TreeNode qe = qStack.peek();
            if (pe.val == qe.val) {
                res = pStack.peek();
                pStack.pop();
                qStack.pop();
                return res;
            }
 
            if (pMap.get(pe) > qMap.get(qe)) {
                pStack.pop();
            } else if (pMap.get(pe) < qMap.get(qe)) {
                qStack.pop();
            } else {
                pStack.pop();
                qStack.pop();
            }
        }
        return null;
    }
 
    void dfs(TreeNode node, int val, Stack<TreeNode> stack, Map<TreeNode, Integer> map, int l) {
        if (node == null) {
            return;
        }
        if (node.val == val) {
            stack.push(node);
            map.put(node, l);
        } else if (node.val > val) {
            stack.push(node);
            map.put(node, l);
            dfs(node.left, val, stack, map, l + 1);
        } else {
            stack.push(node);
            map.put(node, l);
            dfs(node.right, val, stack, map, l + 1);
        }
    }
}
 
// 解决方案(2)
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(p.val > q.val) { // 保证 p.val < q.val
            TreeNode tmp = p;
            p = q;
            q = tmp;
        }
        while(root != null) {
            if(root.val < p.val) // p,q 都在 root 的右子树中
                root = root.right; // 遍历至右子节点
            else if(root.val > q.val) // p,q 都在 root 的左子树中
                root = root.left; // 遍历至左子节点
            else break;
        }
        return root;
    }
}
 
// 解决方案(3)
class Solution {
    public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
        if(root.val < p.val && root.val < q.val)
            return lowestCommonAncestor(root.right, p, q);
        if(root.val > p.val && root.val > q.val)
            return lowestCommonAncestor(root.left, p, q);
        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) {}
 * };
 */

// 解决方案(1)
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(p->val > q->val)
            swap(p, q);
        while(root != nullptr) {
            if(root->val < p->val) // p,q 都在 root 的右子树中
                root = root->right; // 遍历至右子节点
            else if(root->val > q->val) // p,q 都在 root 的左子树中
                root = root->left; // 遍历至左子节点
            else break;
        }
        return root;
    }
};
 
// 解决方案(2)
class Solution {
public:
    TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
        if(root->val < p->val && root->val < q->val)
            return lowestCommonAncestor(root->right, p, q);
        if(root->val > p->val && root->val > q->val)
            return lowestCommonAncestor(root->left, p, q);
        return root;
    }
};