530.二叉搜索树的最小绝对差 501.二叉搜索树中的众数 236. 二叉树的最近公共祖先

530.二叉搜索树的最小绝对差

二叉搜索树中序遍历

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int res = Integer.MAX_VALUE;
    TreeNode pre = null;
    public int getMinimumDifference(TreeNode root) {
        f(root);
        return res;

    }
    void f(TreeNode root){
        if(root == null) return;
        f(root.left);
        if(pre != null){
            res = Math.min(res,root.val - pre.val);
        }
        pre = root;
        f(root.right);
    }
}

迭代法：

class Solution {
    TreeNode pre;
    Stack<TreeNode> stack;
    public int getMinimumDifference(TreeNode root) {
        if (root == null) return 0;
        stack = new Stack<>();
        TreeNode cur = root;
        int result = Integer.MAX_VALUE;
        while (cur != null || !stack.isEmpty()) {
            if (cur != null) {
                stack.push(cur); // 将访问的节点放进栈
                cur = cur.left; // 左
            }else {
                cur = stack.pop(); 
                if (pre != null) { // 中
                    result = Math.min(result, cur.val - pre.val);
                }
                pre = cur;
                cur = cur.right; // 右
            }
        }
        return result;
    }
}

501.二叉搜索树中的众数

和上一题思路一致，计算众数时需要点技巧，不使用额外空间。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    List<Integer> list;
    int maxCount = 0;
    int count = 0;
    TreeNode pre = null;
    public int[] findMode(TreeNode root) {
        list = new ArrayList<>();
        f(root);
        int[] res = new int[list.size()];
        for(int i = 0;i < list.size();i++){
            res[i] = list.get(i);
        }
        return res;
    }
        void f(TreeNode root){
        if(root == null) return;
        f(root.left);
        if(pre == null || pre.val != root.val){
            count = 1;
        }else{
            count++;
        }
        if(count > maxCount){
            maxCount = count;
            list.clear();
            list.add(root.val);
        }else if(count == maxCount){
            list.add(root.val);
        }
        pre = root;
        f(root.right);
    }
}

236. 二叉树的最近公共祖先

/**
 * 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) {
       if (root == null || root == p || root == q) { // 递归结束条件
            return root;
        }

        // 后序遍历
        TreeNode left = lowestCommonAncestor(root.left, p, q);
        TreeNode right = lowestCommonAncestor(root.right, p, q);

        if(left == null && right == null) { // 若未找到节点 p 或 q
            return null;
        }else if(left == null && right != null) { // 若找到一个节点
            return right;
        }else if(left != null && right == null) { // 若找到一个节点
            return left;
        }else { // 若找到两个节点
            return root;
        }
    }
}