8000 从前序与中序遍历序列构造二叉树 · Issue #40 · louzhedong/blog · GitHub
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从前序与中序遍历序列构造二叉树 #40
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@louzhedong

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@louzhedong

习题

出处 LeetCode 算法第105题

根据一棵树的前序遍历与中序遍历构造二叉树。

注意:
你可以假设树中没有重复的元素。

例如,给出

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]

返回如下的二叉树:

    3
   / \
  9  20
    /  \
   15   7

思路

按照思路递归遍历数组,前序遍历的第一个值为树的根,其在中序遍历中对应的值的前面的数都在其左子树上

解答

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {number[]} preorder
 * @param {number[]} inorder
 * @return {TreeNode}
 */
function helper(preorder, inorder) {
  if (preorder.length == 0 || inorder.length == 0) {
    return null;
  }
  var rootVal = preorder[0];
  var index = inorder.indexOf(rootVal);
  var perorderLeft = preorder.slice(1, index + 1);
  var perorderRight = preorder.slice(index + 1);
  var InorderLeft = inorder.slice(0, index);
  var InorderRight = inorder.slice(index + 1);
  var root = new TreeNode(rootVal);
  if (perorderLeft.length > 0) {
    root.left = buildTree(perorderLeft, InorderLeft);
  }
  if (perorderRight.length > 0) {
    root.right = buildTree(perorderRight, InorderRight);
  }
  return root;
}

var buildTree = function (preorder, inorder) {
  if (preorder.length == 0 || inorder.length == 0) {
    return [];
  }
  return helper(preorder, inorder);
};

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