106. Construct Binary Tree from Inorder and Postorder Traversal

题目：

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

链接：  http://leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

题解：

和上一题一样，不过这次的root是从post order的后面来找。

Time Complexity - O(n)， Space Complexity - O(n)。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || postorder == null || inorder.length != postorder.length || inorder.length == 0) 
            return null;
        return buildTree(postorder, 0, postorder.length - 1, inorder, 0, inorder.length - 1);
    }
    
    private TreeNode buildTree(int[] postorder, int postLo, int postHi, int[] inorder, int inLo, int inHi) {
        if(postLo > postHi || inLo > inHi)
            return null;
        TreeNode root = new TreeNode(postorder[postHi]);
        int rootIndex = 0;
        for(int i = inLo; i <= inHi; i++) {
            if(inorder[i] == root.val) {
                rootIndex = i;
                break;
            }
        }
        int leftTreeLen = rootIndex - inLo;
        root.left = buildTree(postorder, postLo, postLo + leftTreeLen - 1, inorder, inLo, rootIndex - 1);
        root.right = buildTree(postorder, postLo + leftTreeLen, postHi - 1, inorder, rootIndex + 1, inHi);
        return root;
    }
}

二刷:

还是使用了一刷的办法，要注意递归的时候postorder的左子树范围是 [postLo, postLo + leftTreeLen - 1]， 右子树是[postLo+ leftTreeLen, postHi - 1]。

可以使用HashMap保存inorder的key, value，这样在递归调用辅助方法时可以O(1)拿到rootVal，就不用顺序搜索了。

Java:

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || postorder == null || inorder.length != postorder.length || inorder.length == 0) return null;
        return buildTree(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
    }
    
    private TreeNode buildTree(int[] inorder, int inLo, int inHi, int[] postorder, int postLo, int postHi) {
        if (inLo > inHi || postLo > postHi) return null;
        int rootVal = postorder[postHi];
        int rootIndexAtInorder = inLo;
        while (rootIndexAtInorder <= inHi) {
            if (inorder[rootIndexAtInorder] == rootVal) break;
            rootIndexAtInorder++;
        }
        int leftTreeLen = rootIndexAtInorder - inLo;
        TreeNode root = new TreeNode(rootVal);
        root.left = buildTree(inorder, inLo, rootIndexAtInorder - 1, postorder, postLo, postLo + leftTreeLen - 1);
        root.right = buildTree(inorder, rootIndexAtInorder + 1, inHi, postorder, postLo + leftTreeLen, postHi - 1);
        return root;
    }
}

Reference:

https://leetcode.com/discuss/10961/my-recursive-java-code-with-o-n-time-and-o-n-space