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98. Validate Binary Search Tree

时间:2022-11-19 18:22:10浏览次数:39  
标签:TreeNode helper Tree tn 98 return Validate null root

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.
Solution 1: //相当于中序遍历,先加入中间节点,再加入左节点,再加入右节点 class Solution {     public boolean isValidBST(TreeNode root) {         if (root == null)             return true;         Stack<TreeNode> s = new Stack<>();         TreeNode pre = null;         s.push(root);         while (!s.isEmpty()) {             TreeNode tn = s.peek();             while (tn != null) {                 tn = tn.left;                                 s.push(tn);                                         }             s.pop();             if (!s.isEmpty()) {                 tn = s.pop();                 if (pre != null && tn.val <= pre.val)                    return false;                 pre = tn;                             s.push(tn.right);             }                     }         return true;     } }   Solution 2://必须是Long类型,例如只有有个2^31-1节点,如果把Long改成Integer,则会报错 class Solution {     Integer prev = null;     boolean isValid = true;     public boolean isValidBST(TreeNode root) {         helper(root);         return isValid;     }          private void helper(TreeNode root) {         if (root == null)             return;         if (isValid == false)             return;         helper(root.left);         if (prev != null) {             if (root.val <= prev)                 isValid = false;         }         prev = root.val;         helper(root.right);                    } }

标签:TreeNode,helper,Tree,tn,98,return,Validate,null,root
From: https://www.cnblogs.com/MarkLeeBYR/p/16906689.html

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