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;    }}