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Total Accepted: 31041 Total Submissions: 115870
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：
修改下上题代码解决，JAVA实现如下：
static public TreeNode buildTree(int[] inorder, int[] postorder) {
return buildTree(postorder, inorder, 0, postorder.length-1, 0, inorder.length-1);
}
static public TreeNode buildTree(int[] postorder, int[] inorder, int pBegin, int pEnd, int iBegin, int iEnd){
if(pBegin>pEnd)
return null;
TreeNode root = new TreeNode(postorder[pEnd]);
int i = iBegin;
for(;i<iEnd;i++)
if(inorder[i]==root.val)
break;
int lenLeft = i-iBegin;
root.left = buildTree(postorder, inorder, pBegin, pBegin+lenLeft-1, iBegin, i-1);
root.right = buildTree(postorder, inorder, pBegin+lenLeft, pEnd-1, i+1, iEnd);
return root;
}
Java for LeetCode 106 Construct Binary Tree from Inorder and Postorder Traversal
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原文地址：http://www.cnblogs.com/tonyluis/p/4524684.html
