README_EN.md

04.05. Legal Binary Search Tree

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Description

Implement a function to check if a binary tree is a binary search tree.

Example 1:

Input:

    2

   / \

  1   3

Output: true

Example 2:

Input:

    5

   / \

  1   4

     / \

    3   6

Output: false

Explanation: Input: [5,1,4,null,null,3,6].

     the value of root node is 5, but its right child has value 4.

Solutions

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    res, t = True, None
    def isValidBST(self, root: TreeNode) -> bool:
        self.isValid(root)
        return self.res

    def isValid(self, root):
        if not root:
            return
        self.isValid(root.left)
        if self.t is None or self.t < root.val:
            self.t = root.val
        else:
            self.res = False
            return
        self.isValid(root.right)

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    private boolean res = true;
    private Integer t = null;
    public boolean isValidBST(TreeNode root) {
        isValid(root);
        return res;
    }

    private void isValid(TreeNode root) {
        if (root == null) {
            return;
        }
        isValid(root.left);
        if (t == null || t < root.val) {
            t = root.val;
        } else {
            res = false;
            return;
        }
        isValid(root.right);
    }
}

Go

• Non-recursive in-order traversal

func isValidBST(root *TreeNode) bool {
	stack := make([]*TreeNode, 0)
	var prev *TreeNode = nil
	node := root
	for len(stack) > 0 || node != nil {
		for node != nil {
			stack = append(stack, node)
			node = node.Left
		}
		node = stack[len(stack)-1]
		stack = stack[:len(stack)-1]
		if prev == nil || node.Val > prev.Val {
			prev = node
		} else {
			return false
		}
		node = node.Right
	}
	return true
}

• Use upper bound and lower bound to check

func isValidBST(root *TreeNode) bool {
	return check(root, math.MinInt64, math.MaxInt64)
}

func check(node *TreeNode, lower, upper int) bool {
	if node == nil {
		return true
	}
	if node.Val <= lower || node.Val >= upper {
		return false
	}
	return check(node.Left, lower, node.Val) && check(node.Right, node.Val, upper)
}

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