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 Input: [2,1,3] Output: trueExample 2:
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.
保证为二叉搜索树,要求左节点小于当前节点,且当前节点小于右节点。并且所有的左节点数都要小于一开始的节点数。
public boolean isValidBST(TreeNode root) { return isValidBST(root,null,null); } public boolean isValidBST(TreeNode root,Integer low,Integer high) { if (root == null) { return true; } int val = root.val; if(low!=null && val <= low) { return false; } if(high!=null && high <= val) { return false; } if(!isValidBST(root.left,low,val)) { return false; } if(!isValidBST(root.right,val,high)) { return false; }
return true; }