【LeetCode】Construct Binary Tree from Inorder and Postorder Traversal 解题报告

标签：二叉树   中序遍历   后序遍历   分治   

【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.

题意：根据二叉树中序遍历和后序遍历的结果，构造该二叉树。

首先明确一下，中序遍历顺序：left - root - right，后序遍历顺序：left  - right - root。

很显然，后序遍历的最后一个节点就是该二叉树的根节点。

找到根节点在中序遍历数组中的位置，那么其之前的元素都是根节点的左支树，其之后的元素都是根节点的右支数。

这样递归左右两个子数组，分别找出左右子树的根节点。

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int[] inorder;
    int[] postorder;
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder.length < 1 || postorder.length  < 1) return null;
        this.inorder = inorder;
        this.postorder = postorder;
        return getRoot(0, inorder.length - 1, inorder.length - 1);
    }
    
    // sub-tree ranges from begin to end in inorder[], it's root is postorder[rootpos]
    public TreeNode getRoot(int begin, int end, int rootpos) {
        if (begin > end) return null;
        TreeNode root = new TreeNode(postorder[rootpos]);
        int i;
        for (i = begin; i <= end; i++) {
            if (inorder[i] == postorder[rootpos]) break;
        }
        // left sub-tree has (i - begin) nodes, right sub-tree has (end - i) nodes
        root.left = getRoot(begin, i - 1, rootpos - 1 - (end - i));
        root.right = getRoot(i + 1, end, rootpos - 1);
        return root;
    }
}

【LeetCode】Construct Binary Tree from Inorder and Postorder Traversal 解题报告

标签：二叉树   中序遍历   后序遍历   分治   

原文地址：http://blog.csdn.net/ljiabin/article/details/41541817