判断一个数组是否是二叉搜索树的后序遍历(java)

分析:

代码:package Tree;{(int[] sequence,int low,int high){if(sequence == null || high-low <0){return false;}int root = sequence[high];int i;//在二叉搜索树左子数的结点小于根结点for(i=low;i<high;i++){//如果值大于根结点,停止,说明这个下标开始为右子树if(sequence[i]>root){break;}}(int j=i; j < high; j++) {if (sequence[j]<root) {return false;}}//判断左子树是不是二叉搜索数boolean left = true;if(i>0){left = verifySequenceBST(sequence,0,i-1);}//判断右子树是不是二叉搜索数boolean right = true;if(i<high){right = verifySequenceBST(sequence,i+1,high);}return (left&&right);}(String[] args) {int[] arr = {5,7,6,9,11,10,8};boolean flag =verifySequenceBST(arr, 0, arr.length-1);System.out.println(flag);int[] arr1 = {7,4,6,5};boolean flag1 = verifySequenceBST(arr1, 0, arr1.length-1);System.out.println(flag1);}}

运行结果:

truefalse

,偶尔为街头独特的风景驻足,

判断一个数组是否是二叉搜索树的后序遍历(java)

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