leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal-递归

原题链接：106. Construct Binary Tree from Inorder and Postorder Traversal相似博文：105. Construct Binary Tree from Preorder and Inorder Traversal | Java最短代码实现

【思路-Java】

本题考查根据中序遍历和后序遍历构造二叉树。其基本思路，就是找到后序遍历的最后一个数的值，然后在中序遍历中找到这个值。以这个值作为根，将中序遍历划分为左、右子树序列——前边的序列为左子树的中序遍历，后边的序列为右子树的中序遍历。照此递归左右子树：

    public TreeNode buildTree(int[] inorder, int[] postorder) {        return createTree(inorder, 0, postorder.length - 1, postorder,                postorder.length - 1);    }    private TreeNode createTree(int[] inorder, int inBeg, int inEnd,            int[] postorder, int postIndex) {        if (postIndex < 0) return null;        int mid = 0;        for (int i = inBeg; i <= inEnd; i++)            if (inorder[i] == postorder[postIndex]) {                mid = i;                break;            }        TreeNode root = new TreeNode(inorder[mid]);        if (mid - inBeg > 0)  //判断左子树节点数是否大于0。下面的postIndex - (inEnd - mid) - 1代表左子树的根节点在postorder中的下标            root.left = createTree(inorder, inBeg, mid - 1, postorder, postIndex - (inEnd - mid) - 1);        if (inEnd - mid > 0)  //判断右子树节点数是否大于0。下面的<span style="font-family: Arial, Helvetica, sans-serif;">postIndex - 1代表右子树的根节点在postorder中的下标</span>            root.right = createTree(inorder, mid + 1, inEnd, postorder, postIndex - 1);        return root;    }

202 / 202 test cases passed. Runtime: 21 ms  Your runtime beats 17.86% of javasubmissions.

【优化】

其实对于上面的思路，在搜寻时，我们用 hashmap 存储 inorder 的值和下标，这样效率会提高很多：

public class Solution {    private HashMap<Integer, Integer> hm = new HashMap<>();    public TreeNode buildTree(int[] inorder, int[] postorder) {        for(int i = 0; i < inorder.length; i++) hm.put(inorder[i], i);        return createTree(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1);    }    private TreeNode createTree(int[] inorder,int istart,int iend,            int[] postorder, int pstart, int pend) {        if(pstart > pend) return null;        int imid = hm.get(postorder[pend]);        TreeNode root = new TreeNode(postorder[pend]);        root.left = createTree(inorder, istart, imid-1, postorder, pstart, pend-1-(iend-imid));        root.right = createTree(inorder, imid+1, iend, postorder, pstart+imid-istart, pend-1);        return root;    }}

202 / 202 test cases passed. Runtime: 5 ms  Your runtime beats 76.10% of javasubmissions.