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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.
題解:BST valid 的充分必要條件是它的中序遍曆是一個有序序列。
遞迴實現樹的中序遍曆,用私人變數lastVal記錄上一個遍曆的節點的值。在一次遞迴,首先遞迴判斷左子樹是否是BST,並且更新lastVal,然後將root的值跟lastVal比較,看root的值是否大於lastVal;然後遞迴判斷右子樹是否是BST。
代碼如下:
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */10 public class Solution {11 private int lastVal = Integer.MIN_VALUE;12 public boolean isValidBST(TreeNode root) {13 if(root == null)14 return true;15 16 if(!isValidBST(root.left))17 return false;18 19 if(root.val <= lastVal)20 return false;21 lastVal = root.val;22 if(!isValidBST(root.right))23 return false;24 return true;25 }26 }
題目的關鍵點是lastVal更新的時機和與root比較的時機。