106. Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree.
Note: You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
if(inorder==null||postorder==null||inorder.length==0||postorder.length==0){
return null;
}
TreeNode treeNode = new TreeNode(postorder[postorder.length-1]);
for(int i=0;i<inorder.length;i++){
if(inorder[i]==postorder[postorder.length-1]){
treeNode.left = buildTree(
Arrays.copyOfRange(inorder, 0 , i),
Arrays.copyOfRange(postorder, 0, i)
);
treeNode.right = buildTree(
Arrays.copyOfRange(inorder, i+1, inorder.length),
Arrays.copyOfRange(postorder, i, postorder.length-1)
);
}
}
return treeNode;
}
}注意数组复制 Arrays.copyOfRange(array, 0, i);