LeetCode题解：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.

题解：通过中序遍历和后序遍历还原二叉树

解决思路：首先要明确一点，对于后序遍历的结果，如果一个元素所在的位置为i，若在中序遍历的i-1位置的元素为该元素的根结点，说明该元素就是所在子树的右儿子（且没有子树），否则存在右子树。左子树倒没什么特别的

代码：

public class Solution {
private int inLen;
private int postLen;

public TreeNode buildTree(int[] inorder, int[] postorder) {
inLen = inorder.length;
postLen = postorder.length;

return buildTree(inorder, postorder, null);
}

private TreeNode buildTree(int[] inorder, int[] postorder, TreeNode root){
if(postLen < 0){
return null;
}

TreeNode node = new TreeNode(postorder[postLen--]);
if(inorder[postLen] != node.val){
node.right = buildTree(inorder, postorder, node);
}

inLen--;

if((root == null) || inorder[inLen] != root.val){
node.left = buildTree(inorder, postorder, node);
}

return node;
}
}