剑指offer,二叉搜索树的后序遍历序列

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输入一个整数数组，判断该数组是不是某二叉搜索树的后序遍历的结果。如果是则输出Yes,否则输出No。假设输入的数组的任意两个数字都互不相同。

解题思路：根据BST后序的特点，最后节点为跟节点，前面都小，后面都大，然后找到分割点，在分别判断两边是否满足。

public class Solution {    public boolean VerifySquenceOfBST(int [] sequence) {        if(sequence == null || sequence.length == 0) return false;        return verify(sequence, 0, sequence.length-1);    }    public boolean verify(int[] sequence, int start, int end){        if(sequence == null || sequence.length == 0) return false;        if(start > end) return true;        int root = sequence[end];        int i = start;        for(; i < end; i++){            if(sequence[i] > root)                break;        }        int j = i;        for(;j < end; j++){            if(sequence[j] < root) return false;        }        boolean left = true;        left = verify(sequence, start, i-1);        boolean right = true;        right = verify(sequence, i, end-1);        return left && right;    }}

public class Solution {    public boolean VerifySquenceOfBST(int [] sequence) {        if(sequence == null || sequence.length == 0) return false;        return verify(sequence, 0, sequence.length-1);    }    public boolean verify(int[] sequence, int start, int end){        if(start > end) return true;        int pivot = end;        int root = sequence[end];        for(int i = start; i < end; i++){            if(sequence[i] > root){                pivot = i;                break;            }        }        for(int i = pivot; i < end; i++){            if(sequence[i] < root) return false;        }        return verify(sequence, start, pivot-1) && verify(sequence, pivot, end-1);    }}

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