Description

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

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

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

tree

Solution

DFS

注意postorder的指針不要算錯(cuò)了。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return buildTreeRecur(inorder, 0, inorder.length - 1,
                             postorder, 0);
    }
    
    public TreeNode buildTreeRecur(int[] inorder, int is, int ie,
                                  int[] postorder, int ps) {
        if (is > ie) {
            return null;
        }
        
        TreeNode root = new TreeNode(postorder[ps + ie - is]);
        int index = search(inorder, is, ie, root.val);
        root.left = buildTreeRecur(inorder, is, index - 1
                                  ,postorder, ps);
        root.right = buildTreeRecur(inorder, index + 1, ie
                                   ,postorder, ps + index - is);
        
        return root;
    }
    
    public int search(int[] nums, int start, int end, int val) {
        while (start <= end && nums[start] != val) {
            ++start;
        }
        
        return start;
    }
}