问题描述：

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

基本思路：

与前一篇《Construct Binary Tree from Preorder and Inorder Traversal 》类似，找出规律进行递归求解。

代码：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {  //java
public TreeNode buildSubTree(int [] preorder, int pbegin, int pend, int [] inorder, int ibegin, int iend){
		TreeNode root = null;
		if(pbegin > pend || ibegin > iend)
			return root;
		if(ibegin == iend){
			TreeNode tmp = new TreeNode(inorder[iend]);
			root = tmp;
			return root;
		}
		
		int value = inorder[iend];
		int pos = 0;
		
		for(int i = pbegin; i <=pend; i++){
			if(value == preorder[i]){
				pos = i;
				break;
			}
		}
		
		int lnum = pos - pbegin;
		int rnum = pend - pos;
		TreeNode ltreeNode = buildSubTree(preorder,pbegin,pos-1,inorder,ibegin,ibegin+lnum-1);
		TreeNode rTreeNode = buildSubTree(preorder, pos+1, pend, inorder, ibegin+lnum, iend-1);
		TreeNode tmp = new TreeNode(inorder[iend]);
		tmp.left = ltreeNode;
		tmp.right = rTreeNode;
		root = tmp;
		return root;
	}
	public TreeNode buildTree(int[] preorder, int[] inorder) {
        TreeNode root = null;
		if(preorder.length == 0)
        	return root;
		 root = buildSubTree(preorder , 0 , preorder.length-1, inorder, 0, inorder.length-1);
		 return root;
    }
}

[leetcode]

原文：http://blog.csdn.net/chenlei0630/article/details/41408623