GuilinDev

Lc0098

05 August 2008

98 Validate Binary Search Tree

题目

Given a binary tree, determine if it is a valid binary search tree (BST).

Assume a BST is defined as follows:

  • The left subtree of a node contains only nodes with keys less than the node’s key.
  • The right subtree of a node contains only nodes with keys greater than the node’s key.
  • Both the left and right subtrees must also be binary search trees.

Example 1:

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    2
   / \
  1   3

Input: [2,1,3]
Output: true

Example 2:

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    5
   / \
  1   4
     / \
    3   6

Input: [5,1,4,null,null,3,6]
Output: false
Explanation: The root node's value is 5 but its right child's value is 4.

分析

一棵BST定义为：

  • 节点的左子树中的值要严格小于该节点的值。
  • 节点的右子树中的值要严格大于该节点的值。
  • 左右子树也必须是二叉查找树。
  • 一个节点的树也是二叉查找树。

如果该二叉树的左子树不为空，则左子树上所有节点的值均小于它的根节点的值； 若它的右子树不空，则右子树上所有节点的值均大于它的根节点的值；它的左右子树也为二叉搜索树。

代码

递归前序遍历，每次限定传送参数的范围

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/**
 * 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 isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    private boolean isValidBST(TreeNode node, long min, long max) {
        if (node == null) {
            return true;
        }
        if (node.val <= min || node.val >= max) {
            return false;
        }
        return isValidBST(node.left, min, node.val) && isValidBST(node.right, node.val, max);
    }
}

递归中序遍历，仔细想想为什么中序遍历只需判断节点的值一次

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class Solution {
    long pre = Long.MIN_VALUE;
    public boolean isValidBST(TreeNode root) {
        if (root == null) {
            return true;
        }
        
        // 按照中序遍历，访问左子树到底
        if (!isValidBST(root.left)) {
            return false;
        }
        
        if (root.val <= pre) { // 递归中“归”的部分进行判断
            return false;
        }
        
        pre = root.val;
        
        // 访问右子树
        return isValidBST(root.right);
    }
}

Iterative

1) Queue的BFS

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class Solution {
    Queue<TreeNode> queue = new LinkedList<>();
    Queue<Long> upperList = new LinkedList<>(), lowerList = new LinkedList<>();
        
    public boolean isValidBST(TreeNode root) {
        long lower = Long.MIN_VALUE, upper = Long.MAX_VALUE, val;
        
        update(root, lower, upper);
        
        while (!queue.isEmpty()) {
            // 这个不用size分层了，只需把node往队列里面加就行
            TreeNode node = queue.poll();
            lower = lowerList.poll();
            upper = upperList.poll();
            if (node == null) {
                continue;
                }
            val = node.val;
            if (val <= lower) {
                return false;
                }
            if (val >= upper) {
                return false;
                }
            update(node.left, lower, val);
            update(node.right, val, upper);
        }
        return true;
    }

    void update(TreeNode node, long lower, long upper) {
        queue.offer(node);
        lowerList.offer(lower);
        upperList.offer(upper);
    }
}

2) Stack迭代实现前序遍历

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class Solution {
    Stack<TreeNode> st = new Stack<>();
    Stack<Long> upperList = new Stack<>(), 
        lowerList = new Stack<>();
        
    public boolean isValidBST(TreeNode root) {
        long lower = Long.MIN_VALUE, upper = Long.MAX_VALUE, val;
        update(root, lower, upper);
        while (!st.empty()) {
            root = st.pop();
            lower = lowerList.pop();
            upper = upperList.pop();
            if (root == null) continue;
            val = (long)root.val;
            if (val <= lower) return false;
            if (val >= upper) return false;
            update(root.right, val, upper);
            update(root.left, lower, val);
        }
        return true;
    }

    void update(TreeNode node, long lower, long upper) {
        st.push(node);
        lowerList.push(lower);
        upperList.push(upper);
    }
}
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