Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
思路:经典的题目。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
TreeNode build(int[] preorder, int[] inorder, int l, int r, int len) {
TreeNode root = null;
if (len < 1) return root;
root = new TreeNode(preorder[l]);
int index = 0;
for (int i = 0; i < len; i++)
if (inorder[r+i] == preorder[l]) {
index = i;
break;
}
root.left = build(preorder, inorder, l+1, r, index);
root.right = build(preorder, inorder, l+1+index, r+1+index, len-1-index);
return root;
}
public TreeNode buildTree(int[] preorder, int[] inorder) {
int n = preorder.length;
return build(preorder, inorder, 0, 0, n);
}
}
LeetCode Construct Binary Tree from Preorder and Inorder Traversal
原文:http://blog.csdn.net/u011345136/article/details/45437741