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98. 验证二叉搜索树

98. 验证二叉搜索树

作者: justonemoretry | 来源:发表于2021-08-23 20:48 被阅读0次
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解法

递归解法

/**
 * 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) {
        return dfs(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }

    /**
     * 递归遍历,判断每个节点是否符合范围   
     */
    private boolean dfs(TreeNode node, long floor, long ceil) {
        if (node == null) {
            return true;
        }
        int val = node.val;
        if (val <= floor || val >= ceil) {
            return false;
        }
        // 左节点小于当前节点的值
        // 右节点大于当前节点的值
        return dfs(node.left, floor, val) && dfs(node.right, val, ceil);
    }
}

迭代解法

/**
 * 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) {
        long pre = Long.MIN_VALUE;
        Stack<TreeNode> s = new Stack<>();
        // 中序遍历,当前遍历节点大于之前访问到节点的值
        while (!s.isEmpty() || root != null) {
            while (root != null) {
                s.push(root);
                root = root.left;
            }
            TreeNode node = s.pop();
            if (node.val <= pre) {
                return false;
            }
            pre = node.val;
            root = node.right;
        }
        return true;
    }
}

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