Given preorder and inorder traversal of a tree, construct the binary tree.
?
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public TreeNode buildTree(int[] preorder, int[] inorder) {
return buildTree(preorder, 0, preorder.length, inorder, 0, inorder.length);
}
private TreeNode buildTree(int[] preorder, int i, int length,
int[] inorder, int j, int length2) {
if (i>length-1 || j>length2-1) {
return null;
}
int tmp = preorder[i];
int index = 0;
for (int k = j; k < length2; k++) {
if (inorder[k] == tmp) {
index = k;
break;
}
}
int len = index-j;
TreeNode root = new TreeNode(tmp);
root.left = buildTree(preorder, i+1, i+len+1, inorder, j, index);
root.right = buildTree(preorder, i+1+len, length, inorder, index+1, length2);
return root;
}
}
?
Construct Binary Tree from Preorder and Inorder Traversal
原文：http://hcx2013.iteye.com/blog/2237966