Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
思路:还是简单的递归构造。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private TreeNode build(int[] inorder, int[] postorder, int l, int r, int len) {
if (len <= 0) return null;
int rootVal = postorder[r+len-1];
TreeNode root = new TreeNode(rootVal);
int index = 0;
for (int i = 0; i < len; i++)
if (inorder[l+i] == rootVal) {
index = i;
break;
}
root.left = build(inorder, postorder, l, r, index);
root.right = build(inorder, postorder, l+index+1, r+index, len-1-index);
return root;
}
public TreeNode buildTree(int[] inorder, int[] postorder) {
return build(inorder, postorder, 0, 0, inorder.length);
}
}LeetCode Construct Binary Tree from Inorder and Postorder Traversal
原文:http://blog.csdn.net/u011345136/article/details/45456883