leetcode106

基础题:根据二叉树的中缀和后缀求前缀
思路,递归分治即可

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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
int len = postorder.length;
if (len==0){
return null;
}
int x = postorder[len-1];
TreeNode root = new TreeNode(x);
if (len==1){
return root;
}
int index = -1;
for(int i=0;i<len;i++){
if (inorder[i]==x){
index = i;
break;
}
}

if (index>0){
int[] leftInorder = new int[index];
int[] leftPostorder = new int[index];
for(int i=0;i<index;i++){
leftInorder[i] = inorder[i];
}
for(int i=0;i<index;i++){
leftPostorder[i] = postorder[i];
}
TreeNode left = buildTree(leftInorder,leftPostorder);
root.left = left;
}

if (len-1-index>0){
int[] leftInorder = new int[len-1-index];
int[] leftPostorder = new int[len-1-index];
for(int i=0;i<len-1-index;i++){
leftInorder[i] = inorder[i+index+1];
}
for(int i=0;i<len-1-index;i++){
leftPostorder[i] = postorder[i+index];
}
TreeNode right = buildTree(leftInorder,leftPostorder);
root.right = right;
}
return root;
}
}

递归的消耗还是大,内存是时间复杂度都比较高,特别是数组复制,但是这样清晰
1.一种优化是,不复制数组,将下标传入即可
2.移除递归,改为while循环