106. 从中序与后序遍历序列构造二叉树（递归版）

/**
 * 106. 从中序与后序遍历序列构造二叉树
 * @author wsq
 * @date 2020/09/25
	根据一棵树的中序遍历与后序遍历构造二叉树。
	注意:
	你可以假设树中没有重复的元素。

	例如，给出
	中序遍历 inorder = [9,3,15,20,7]
	后序遍历 postorder = [9,15,7,20,3]
	
	返回如下的二叉树：
	    3
	   / \
	  9  20
	    /  \
	   15   7
	
	链接：https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal
 */
package notsubmit;

public class MidPosBuildTree {
	/**
	 *	根据后序遍历与中序遍历的特点
	 *	采用递归的方式构建二叉树
	 *	以后会尝试使用迭代的方式实现树的构建
	 * @param inorder
	 * @param postorder
	 * @return
	 */
    public TreeNode buildTree(int[] inorder, int[] postorder) {
    	int inLen = inorder.length;
    	int posLen = postorder.length;
    	if(inLen == 0 || posLen == 0 || inLen != posLen) {
    		return null;
    	}
    	// 递归建树
    	TreeNode root = reBuildTree(inorder, postorder, 0, inLen - 1, 0, posLen - 1);
    	return root;
    }
    
    private TreeNode reBuildTree(int[] inorder, int[] postorder, 
    							 int inStart, int inEnd, 
    							 int postStart, int postEnd) {
		// TODO Auto-generated method stub
    	if(inStart > inEnd || postStart > postEnd) {
    		return null;
    	}
    	int rootVal = postorder[postEnd];
    	TreeNode rootNode = new TreeNode(postorder[postEnd]);
    	
    	if(postStart == postEnd) {
    		return rootNode;
    	}
    	
    	int inPos = findInorder(inorder, inStart, inEnd, rootVal);
    	int leftTreeNum = inPos - inStart;

    	rootNode.left = reBuildTree(inorder, postorder, inStart, inPos - 1, postStart, postStart + leftTreeNum - 1);
    	rootNode.right = reBuildTree(inorder, postorder, inPos + 1, inEnd, postStart + leftTreeNum, postEnd - 1);
		return rootNode;
	}

	private int findInorder(int[] inorder, int inStart, int inEnd, int rootVal) {
		// TODO Auto-generated method stub
		for(int i = inStart; i <= inEnd; ++i) {
			if(inorder[i] == rootVal) {
				return i;
			}
		}
		return -1;
	}
	
	public static void main(String[] args) {
		int[] inorder = {2, 1};
		int[] postorder = {2, 1};
		MidPosBuildTree bt = new MidPosBuildTree();
		TreeNode head = bt.buildTree(inorder, postorder);
	}

}