Construct Binary Tree from Inorder and Postorder Traversal

最新推荐文章于 2021-12-13 22:59:20 发布

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最新推荐文章于 2021-12-13 22:59:20 发布

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分类专栏： lintcode interview question 文章标签： Binary Tree

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本文介绍了一种通过给定的中序遍历和后序遍历序列来构建二叉树的方法。该方法首先检查输入序列的有效性，然后递归地构建二叉树结构。通过查找后序遍历中的根节点，在中序遍历中确定左右子树的范围，从而完成树的构建。

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

Notice

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

Given inorder [1,2,3] and postorder [1,3,2], return a tree:

  2
 / \
1   3

/**
 * 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.length == 0 || postorder.length == 0) {
            return null;
        }
        return construct(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
    }
    
    private TreeNode construct(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) {
        if(inStart > inEnd || postStart > postEnd) {
            return null;
        }
        TreeNode root = new TreeNode(postorder[postEnd]);
        int position = 0;
        for(int i = 0; i < inorder.length; i++) {
            if(inorder[i] == postorder[postEnd]) {
                position = i;
            }
        }
        int leftLength = position - inStart;
        root.left = construct(inorder, inStart, position - 1, postorder, postStart, postStart + leftLength - 1);//be careful about the index
        root.right = construct(inorder, position + 1, inEnd, postorder, postStart + leftLength, postEnd - 1);
        return root;
    }
}