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.

ref: http://fisherlei.blogspot.com/2013/01/leetcode-construct-binary-tree-from.html

注意 preorder根节点的下标是 pStart, 不要写成 0

 注意 preorder 注意左子树和右子树开始的index

左子树是 pStart+1 + (rootIndex-1 - iStart) // inorder中 左子树的长度

右子树是 pStart+1 + (rootIndex-1 - iStart) +1

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public  TreeNode buildTree(int[] preorder, int[] inorder) {
        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;
        // 如果左子树不为空
        if (leftLen > 0) {
            root.left = constructTree(preorder, preStart + 1, preStart
                    + leftLen, inorder, inStart, i - 1);
        }
        // 如果右子树不为空
        if (inEnd > i) {
            root.right = constructTree(preorder, preStart + leftLen + 1,
                    preEnd, inorder, i + 1, inEnd);
        }
        return root;
    }

}