leetcode -- Construct Binary Tree from Preorder and Inorder Traversal

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

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

[解题思路]

由前序遍历知第一个节点是根节点，根据此节点值去中序遍历集合中找该root值所处位置，

该位置之前的数属于左子树，之后的属于右子树，即找到左子树和右子树所处的子序列，接下来就可以用递归来完成

递归的终止条件：preorder中仅有一个元素且preorder中的元素和inorder中元素相同

/**
 * @author rgc
 * @date 2013-8-23上午10:07:40
 * @version 1.0
 */
public class Solution {

    public static void main(String[] args) {
        int[] preorder = new int[]{1,2,3};
        int[] inorder = new int[]{1,2,3};
        buildTree(preorder, inorder);
    }

    public static TreeNode buildTree(int[] preorder, int[] inorder) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if (preorder == null || inorder == null) {
            return null;
        }
        int preLen = preorder.length;
        int inLen = inorder.length;
        if (preLen == 0 || inLen == 0) {
            return null;
        }

        return constructTree(preorder, 0, preLen - 1, inorder, 0, inLen - 1);
    }

    public static TreeNode constructTree(int[] preorder, int preStart, int preEnd,
            int[] inorder, int inStart, int inEnd) {
        int rootValue = preorder[preStart];
        TreeNode root = new TreeNode(rootValue);
        root.left = null;
        root.right = null;
        if (preStart == preEnd && preorder[preStart] == inorder[inStart]) {
            return root;
        }

        int i = inStart;
        for (; i <= inEnd; i++) {
            if (rootValue == inorder[i]) {
                break;
            }
        }
        int leftLen = i - inStart;
        // exsit left subtree
        if (leftLen > 0) {
            root.left = constructTree(preorder, preStart + 1, preStart
                    + leftLen, inorder, inStart, i - 1);
        }
        if (inEnd > i) {
            root.right = constructTree(preorder, preStart + leftLen + 1,
                    preEnd, inorder, i + 1, inEnd);
        }
        return root;
    }

}

class TreeNode {
    int val;
    TreeNode left;
    TreeNode right;

    TreeNode(int x) {
        val = x;
    }
}