重建二叉树
题目：
输入某二叉树的前序遍历和中序遍历的结果，请重建出该二叉树。假设输入的前序遍历和中序遍历的结果中都不含重复的数字。例如输入前序遍历序列{1,2,4,7,3,5,6,8}和中序遍历序列{4,7,2,1,5,3,8,6}，则重建二叉树并返回。
思路：在代码中
代码：
public TreeNode reConstructBinaryTree(int[] pre,int[] in) {
TreeNode root = null;
return createTreeBypreAndIn(root, pre, 0, pre.length-1, in, 0, in.length-1);
}
private TreeNode createTreeBypreAndIn(TreeNode root,int[] preOrder, int preStart, int preEnd,int[] inOrder, int inStart, int inEnd) {
//创建结点，先序遍历的第一个便是根节点
root = new TreeNode(preOrder[preStart]);
//在中序中找出根的位置
int _index = inStart;
while (preOrder[preStart] != inOrder[_index]) {
_index++;
}
//假如 _index = inStart 说明没有左结点
if (_index == inStart) {
root.left = null;
} else {
/**
* 递归建立左子树, 则左子树的建树空间变为:
* preStart = preStart+1
* preEnd = preStart + _index -1
* inStart = inStart;
* inEnd = _index-1
*/
root.left = createTreeBypreAndIn(root.left, preOrder, preStart + 1, preStart + _index - 1, inOrder, inStart, _index - 1);
}
//假如 _index = inEnd 说明没有右结点
if (_index == inEnd) {
root.right = null;
} else {
/**
* 递归建立右子树, 则右子树的建树空间变为:
* preStart = preStart + _index - inStart + 1(_index - inStart + 1代表左子树有多少个结点)
* preEnd = preEnd
* inStart = _index + 1
* inEnd = inEnd
*/
root.right = createTreeBypreAndIn(root.right, preOrder, preStart + _index - inStart + 1, preEnd,inOrder, _index + 1, inEnd);
}
return root;
}
