106. 从中序与后序遍历序列构造二叉树

题目

根据一棵树的中序遍历与后序遍历构造二叉树。

注意: 你可以假设树中没有重复的元素。

例如，给出

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]
返回如下的二叉树：

    3
   / \
  9  20
    /  \
   15   7

题解

递归

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {number[]} inorder
 * @param {number[]} postorder
 * @return {TreeNode}
 */
var buildTree = function(inorder, postorder) {
  if (!postorder.length) return null;
  let root = new TreeNode(postorder[postorder.length - 1]);
  let i = 0;
  for (; i < inorder.length; ++i) {
    if (inorder[i] == root.val) {
      break;
    }
  }

  root.left = buildTree(inorder.slice(0, i), postorder.slice(0, i));
  root.right = buildTree(inorder.slice(i + 1), postorder.slice(i, -1));
  return root;
};

迭代

var buildTree = function(inorder, postorder) {
  if (postorder.length == 0) {
    return null;
  }
  const root = new TreeNode(postorder[postorder.length - 1]);
  const stack = [];
  stack.push(root);
  let inorderIndex = inorder.length - 1;
  for (let i = postorder.length - 2; i >= 0; i--) {
    let postorderVal = postorder[i];
    let node = stack[stack.length - 1];
    if (node.val !== inorder[inorderIndex]) {
      node.right = new TreeNode(postorderVal);
      stack.push(node.right);
    } else {
      while (
        stack.length &&
        stack[stack.length - 1].val === inorder[inorderIndex]
      ) {
        node = stack.pop();
        inorderIndex--;
      }
      node.left = new TreeNode(postorderVal);
      stack.push(node.left);
    }
  }
  return root;
};