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;
}
}