Question

Source

• lintcode: (72) Construct Binary Tree from Inorder and Postorder Traversal

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

Example
Given inorder [1,2,3] and postorder [1,3,2], return a tree:
  2
   / \
   1   3
   Note
   You may assume that duplicates do not exist in the tree.

题解

和题 Construct Binary Tree from Preorder and Inorder Traversal 几乎一致，关键在于找到中序遍历中的根节点和左右子树，递归解决。

Java

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *   public int val;
 *   public TreeNode left, right;
 *   public TreeNode(int val) {
 *     this.val = val;
 *     this.left = this.right = null;
 *   }
 * }
 */

public class Solution {
  /**
   *@param inorder : A list of integers that inorder traversal of a tree
   *@param postorder : A list of integers that postorder traversal of a tree
   *@return : Root of a tree
   */
  public TreeNode buildTree(int[] inorder, int[] postorder) {
    if (inorder == null || postorder == null) return null;
    if (inorder.length == 0 || postorder.length == 0) return null;
    if (inorder.length != postorder.length) return null;

    TreeNode root = helper(inorder, 0, inorder.length - 1,
         postorder, 0, postorder.length - 1);
    return root;
  }

  private TreeNode helper(int[] inorder, int instart, int inend,
              int[] postorder, int poststart, int postend) {
    // corner cases
    if (instart > inend || poststart > postend) return null;

    // build root TreeNode
    int root_val = postorder[postend];
    TreeNode root = new TreeNode(root_val);
    // find index of root_val in inorder[]
    int index = findIndex(inorder, instart, inend, root_val);
    // build left subtree
    root.left = helper(inorder, instart, index - 1,
               postorder, poststart, poststart + index - instart - 1);
    // build right subtree
    root.right = helper(inorder, index + 1, inend,
               postorder, poststart + index - instart, postend - 1);
    return root;
  }

  private int findIndex(int[] nums, int start, int end, int target) {
    for (int i = start; i <= end; i++) {
      if (nums[i] == target) return i;
    }
    return -1;
  }
}

源码分析

找根节点的方法作为私有方法，辅助函数需要注意索引范围。

复杂度分析

找根节点近似 O(n)O(n)O(n), 递归遍历整个数组，嵌套找根节点的方法，故总的时间复杂度为 O(n2)O(n^2)O(n2).