leetcode98. Validate Binary Search Tree

题目要求

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:
    2
   / \
  1   3
Binary tree [2,1,3], return true.
Example 2:
    1
   / \
  2   3
Binary tree [1,2,3], return false.

判断一个树是不是二叉查找树。二叉查找树即知足当前节点左子树的值均小于当前节点的值，右子树的值均大于当前节点的值。

思路一：stack dfs

能够看到，对二叉查找树的中序遍历结果应当是一个递增的数组。这里咱们用堆栈的方式实现中序遍历。

public boolean isValidBST(TreeNode root) {
        long currentVal = Long.MIN_VALUE;
        LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
        while(root!=null || !stack.isEmpty()){
            while(root!=null){
                stack.push(root);
                root = root.left;
            }
            root = stack.pop();
            if(root.val<=currentVal) return false;
            currentVal = root.val;
            root = root.right;
        }
        return true;
    }

思路二：递归

咱们能够发现，若是已知当前节点的值val以及取值的上下限upper，lower，那么左子节点的取值范围就是(lower, val),右子节点的取值范围就是(val, upper)。由此出发递归判断，时间复杂度为O(n)由于每一个节点只须要遍历一次。

public boolean isValidBST2(TreeNode root){
        return isValid(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
    
    public boolean isValid(TreeNode treeNode, long lowerBound, long upperBound){
        if(treeNode==null) return true;
        if(treeNode.val>=upperBound || treeNode.val<=lowerBound) return false;
        return isValid(treeNode.left, lowerBound,treeNode.val) && isValid(treeNode.right, treeNode.val, upperBound);
    }

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