LeetCode 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.

解题思路：这题和105的解题思路类似，主要区别在于：将先序遍历换成后续后，根节点将是每部分孩子节点的最后一个节点，所以只需要改变的只是根节点在遍历数组中的下标。首先得到树的根节点后，在中序数组中区分左右孩子节点，然后计算左右孩子节点的个数，找到左右孩子的根节点在后续遍历数组中的下标。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode buildTree(int[] inorder, int[] postorder) {
TreeNode root=null;
if(inorder.length==postorder.length){
if(inorder.length>=0)
root= helper(postorder,postorder.length-1,inorder,0,inorder.length);
}
return root;
}
public TreeNode helper(int[] postorder,int index,int [] inorder,int start,int end){

if(index>= postorder.length || index <0 || start >=end){
return null;
}else{
TreeNode root =new TreeNode(postorder[index]);
int median =0;
for(int i=start;i<end;i++){
if(postorder[index]==inorder[i]){
median = i;
break;
}
}
root.left=helper(postorder,index-end+median,inorder,start,median);
root.right=helper(postorder,index-1,inorder,median+1,end);
return root;
}
}
}