JavaScript 打印二叉树左视图的实现返回不正确的结果
JavaScript implementation of printing left view of binary tree is returning incorrect result
我正在尝试打印在 geeksforgeeks 上 here 看到的二叉树的左视图。由于某种原因它不起作用,我怀疑它与 max_level
有关。结果是 [ 12, 10, 30, 25, 40 ]
,我期待 [12,10,25]
。
JS代码
var Node = function(val) {
this.val = val;
this.left = this.right = null;
};
var leftViewUtil = function(root, level, max, result) {
if (root === null) return;
if (max.level < level) {
max.level = level;
result.arr.push(root.val);
}
leftViewUtil(root.left, ++level, max, result);
leftViewUtil(root.right, ++level, max, result);
};
var leftView = function(root) {
var result = {
arr: []
};
var max_level = {level: 0};
leftViewUtil(root, 1, max_level, result);
return result.arr;
};
root = new Node(12);
root.left = new Node(10);
root.right = new Node(30);
root.right.left = new Node(25);
root.right.right = new Node(40);
var run = function() {
console.log(leftView(root));
};
run();
链接页面代码的区别是
// Recur for left and right subtrees
leftViewUtil(root->left, level+1, max_level);
leftViewUtil(root->right, level+1, max_level);
对
leftViewUtil(root.left, ++level, max, result);
leftViewUtil(root.right, ++level, max, result);
您在这里增加了 level
两次,而您应该将相同的值传递给两个递归调用。适当使用 level+1
,或者在调用之前进行递增:
++level;
leftViewUtil(root.left, level, max, result);
leftViewUtil(root.right, level, max, result);
使用哈希table在几行代码中找到树的左右视图。
right_view(root,num, result) {
if(root == null) {
return 0
}
right_view(root.Left, num+1, result)
right_view(root.Right, num+1, result)
result[num] = root.Value
}
left_view(root,num, result) {
if(root == null) {
return 0
}
left_view(root.Left, num+1, result)
left_view(root.Right, num+1, result)
if(result[num] == undefined) {
result[num] = root.Value
}
}
用根节点调用函数。
right_view_result = {}
right_view(root,1,right_view_result)
console.log(right_view_result)
用根节点调用函数。
left_view_result = {}
left_view(root,1,left_view_result)
console.log(left_view_result)
我正在尝试打印在 geeksforgeeks 上 here 看到的二叉树的左视图。由于某种原因它不起作用,我怀疑它与 max_level
有关。结果是 [ 12, 10, 30, 25, 40 ]
,我期待 [12,10,25]
。
JS代码
var Node = function(val) {
this.val = val;
this.left = this.right = null;
};
var leftViewUtil = function(root, level, max, result) {
if (root === null) return;
if (max.level < level) {
max.level = level;
result.arr.push(root.val);
}
leftViewUtil(root.left, ++level, max, result);
leftViewUtil(root.right, ++level, max, result);
};
var leftView = function(root) {
var result = {
arr: []
};
var max_level = {level: 0};
leftViewUtil(root, 1, max_level, result);
return result.arr;
};
root = new Node(12);
root.left = new Node(10);
root.right = new Node(30);
root.right.left = new Node(25);
root.right.right = new Node(40);
var run = function() {
console.log(leftView(root));
};
run();
链接页面代码的区别是
// Recur for left and right subtrees leftViewUtil(root->left, level+1, max_level); leftViewUtil(root->right, level+1, max_level);
对
leftViewUtil(root.left, ++level, max, result); leftViewUtil(root.right, ++level, max, result);
您在这里增加了 level
两次,而您应该将相同的值传递给两个递归调用。适当使用 level+1
,或者在调用之前进行递增:
++level;
leftViewUtil(root.left, level, max, result);
leftViewUtil(root.right, level, max, result);
使用哈希table在几行代码中找到树的左右视图。
right_view(root,num, result) {
if(root == null) {
return 0
}
right_view(root.Left, num+1, result)
right_view(root.Right, num+1, result)
result[num] = root.Value
}
left_view(root,num, result) {
if(root == null) {
return 0
}
left_view(root.Left, num+1, result)
left_view(root.Right, num+1, result)
if(result[num] == undefined) {
result[num] = root.Value
}
}
用根节点调用函数。
right_view_result = {}
right_view(root,1,right_view_result)
console.log(right_view_result)
用根节点调用函数。
left_view_result = {}
left_view(root,1,left_view_result)
console.log(left_view_result)