leetCode 106.Construct Binary Tree from Inorder and Postorder Traversal (根据中序遍历和后序遍历构造二叉树)

标签：二叉树   leetcode   遍历   dfs   

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

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

代码如下：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
         /**
         * 1.根据后序遍历，先确定根节点
         * 2.然后在中序遍历中查找根节点，确定根节点在中序遍历的位置
         * 3.根据索引位置分割左右子树的前序和中序遍历
         * 4.递归求解根节点的左右子树
         */ 
        if(postorder.length == 0 || inorder.length == 0)
            return null;
        TreeNode root = new TreeNode(postorder[postorder.length-1]);
        int k = 0;
        for(; k < inorder.length; k++){
            if(inorder[k] == postorder[postorder.length-1]){
                break;
            }
        }
        int[] p1 = Arrays.copyOfRange(postorder,0,k);
        int[] q1 = Arrays.copyOfRange(postorder,k,postorder.length-1);
        int[] p2 = Arrays.copyOfRange(inorder,0,k);
        int[] q2 = Arrays.copyOfRange(inorder,k+1,inorder.length);
        
        root.left = buildTree(p2,p1);
        root.right = buildTree(q2,q1);
        return root;
    }
}

leetCode 106.Construct Binary Tree from Inorder and Postorder Traversal (根据中序遍历和后序遍历构造二叉树)

标签：二叉树   leetcode   遍历   dfs   

原文地址：http://blog.csdn.net/xygy8860/article/details/47189651