Construct Binary Tree from Inorder and Postorder Traversal 由中序和后序重建二叉树@LeetCode

同上一题，这一题是由inorder和postorder来确定树，对于postorder，root要从尾部开始找

另外，在从一个order到另外一个order时，要用相对距离来计算！即从第一个order算出dist是多少，然后应用这个dist到第二个order上

package Level4;

import Utility.TreeNode;

/**
 * Construct Binary Tree from Inorder and Postorder Traversal
 * 
 *  Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.
 *
 */
public class S141 {

	public static void main(String[] args) {
		int[] inorder = {1,2,3};
		int[] postorder = {3,2,1};
		
		TreeNode root = buildTree(inorder, postorder);
		root.print();
	}

	public static TreeNode buildTree(int[] inorder, int[] postorder) {
		if(inorder.length == 0){
			return null;
		}
        return rec(inorder, postorder, 0, inorder.length-1, 0, postorder.length-1);
    }
	
	public static TreeNode rec(int[] inorder, int[] postorder, int inStart, int inEnd, int postStart, int postEnd){
		if(postEnd < 0 || postEnd>=postorder.length){
			return null;
		}
		TreeNode root = new TreeNode(postorder[postEnd]);
		int rootIndex;			// rootIndex in inorder[]
		for(rootIndex=0; rootIndex<inorder.length; rootIndex++) {
			if(inorder[rootIndex] == postorder[postEnd]){
				break;
			}
		}
		
		int leftSubTreeLen = rootIndex - inStart;		// inorder中，inStart到rootIndex的距离
		if(rootIndex > inStart){		// 锁定范围，否则会Memory out of limit!
			root.left = rec(inorder, postorder, inStart, rootIndex-1, postStart, postStart+leftSubTreeLen-1);
		}
		if(rootIndex < inEnd){
			root.right = rec(inorder, postorder, rootIndex+1, inEnd, postStart+leftSubTreeLen, postEnd-1);
		}
		
		return root;
	}
}

/**
 * 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[] inorder, int[] postorder) {
        return rec(inorder, postorder, 0, inorder.length-1, 0, postorder.length-1);
    }
    
    public TreeNode rec(int[] inorder, int[] postorder, int inorderLeft, int inorderRight, int postorderLeft, int postorderRight) {
        if(inorderLeft > inorderRight || postorderLeft > postorderRight) {
            return null;
        }
        int root = postorder[postorderRight];
        TreeNode rootNode = new TreeNode(root);
        int inorderRootPos = inorderLeft;
        for(inorderRootPos=inorderLeft; inorderRootPos<=inorderRight; inorderRootPos++) {
            if(inorder[inorderRootPos] == root) {
                break;
            }
        }
        
        int dist = inorderRootPos - inorderLeft;
        rootNode.left = rec(inorder, postorder, inorderLeft, inorderRootPos-1, postorderLeft, postorderLeft+dist-1);
        rootNode.right = rec(inorder, postorder, inorderRootPos+1, inorderRight, postorderLeft+dist, postorderRight-1);
        
        return rootNode;
    }
}