106. Construct Binary Tree from Inorder and Postorder Traversal

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本文介绍了一种算法，用于根据给定的中序和后序遍历序列构建二叉树。通过逆向遍历后序序列，首先构建根节点，然后递归地构建右子树和左子树。文章提供了详细的实现代码和示例。

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106. Construct Binary Tree from Inorder and Postorder Traversal

Medium

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:

Tips: iterate backward postorder. build tree right first, then left.

/**
 * 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) {
        if (null == postorder || postorder.length < 1) {
            return null;
        }
        
        int[] idx = {postorder.length - 1};
        
        return buildTree(postorder, idx, inorder, 0, inorder.length - 1);
    }
    
    private TreeNode buildTree(int[] postorder, int[] idx, int[] inorder, int s, int e) {
        int rIdx = s;
        for (int i = s; i <= e; i++) {
            if (postorder[idx[0]] == inorder[i]) {
                rIdx = i;
                break;
            }
        }

        TreeNode root = new TreeNode(postorder[idx[0]]);
        //right
        if (e - rIdx >= 1) {
            idx[0] -= 1;
            root.right = buildTree(postorder, idx, inorder, rIdx + 1, e);
        }
        // left
        if (rIdx - s >= 1) {
            idx[0] -= 1;
            root.left = buildTree(postorder, idx, inorder, s, rIdx - 1);
        }
        
        return root;   
    }
}