106. Construct Binary Tree from Inorder and Postorder Traversal

Posted 2020-08-14 鱼与海洋

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

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

 

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/**
 * 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) {
        return buildSubTree(0, postorder.length -1, postorder, inorder, 0, postorder.length-1);
    }
    public TreeNode buildSubTree(int postLeft, int postRight, int[] postorder, int[] inorder, int inLeft, int inRight){
        if(postRight < postLeft || inLeft > inRight) return null;
        int mid = inLeft;
        TreeNode root = new TreeNode(postorder[postRight]);
        while(mid <= inRight){
            if(postorder[postRight] == inorder[mid]){
                root.left = buildSubTree(postLeft, postRight+mid-inRight -1 , postorder, inorder, inLeft, mid-1);
                root.right = buildSubTree(postRight+mid-inRight, postRight -1, postorder, inorder, mid+1, inRight);
            }
            mid++;
        }
        return root;
    }
}