🌕 98. 验证二叉搜索树

难度: 🌕

问题描述

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解法 1 - 递归

/**
 * 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 {
    TreeNode pre = null;
    public boolean isValidBST(TreeNode root) {
        // 思路：
        // 中序遍历 - 二叉搜索时有序
        return mySol(root);
    }

    private boolean mySol(TreeNode root) {
        // 递归终止条件
        if(root == null) {
            return true;
        }
        boolean left = mySol(root.left);
        if(!left) {
            return false;
        }
        int cur = root.val;
        if(pre != null && cur <= pre.val) {
            return false;
        }
        pre = root;
        boolean right = mySol(root.right);
        return right;
    }
}

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输出 1

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解法 2 - 迭代

/**
 * 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 {
    public boolean isValidBST(TreeNode root) {
        // 思路：
        // 迭代
        LinkedList<TreeNode> stack = new LinkedList<>();
        TreeNode pre = null;
        TreeNode cur = root;
        while(cur != null || !stack.isEmpty()) {
            while(cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            if(pre != null && cur.val <= pre.val) {
                return false;
            }
            pre = cur;
            cur = cur.right;
        }
        return true;
    }
}

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输出 2