递归遍历二叉树Javascript的所有嵌套子节点

Question

I am playing around with a binary tree. I am trying to use recursion to find all the nested children's values and push all the values into an array. I started with the left tree to see if it works. I tried to call childrenArray() but the console says childrenArray() is not defined. When I ask if (typeof BinaryTree.prototype.childrenArray === 'function') , it returns true . Please teach me and tell me why I wasn't able to execute my code?

var Tree = function(value) {
  if (!(this instanceof Tree)) {
    return new Tree(value);
  }
  this.value = value;
  this.children = [];
};

Tree.prototype.addChild = function(value) {
  var child = new Tree(value);
  this.children.push(child);
};

Tree.prototype.contains = function(value) {
  if (this.value === value) {
    return true;
  } else {
    for (var i = 0; i < this.children.length; i++) {
      if (this.children[i] && this.children[i].contains(value)) {
        return true;
      }
    }
    return false;
  }
};


var BinaryTree = function(value) {
  if (!(this instanceof BinaryTree)) {
    return new BinaryTree(value);
  }
  Tree.call(this, value);
};
BinaryTree.prototype = Object.create(Tree.prototype);

BinaryTree.prototype.addChild = function(value) {
  if (value < this.value) {
    if (this.children[0] === undefined) {
      this.children[0] = new BinaryTree(value);
    }
    this.children[0].addChild(value);
  } else if (value > this.value) {
    if (this.children[1] === undefined) {
      this.children[1] = new BinaryTree(value);
    }
    this.children[1].addChild(value);
  }
};
BinaryTree.prototype.contains = function(value) {
  if (value < this.value) {
    if (this.children[0] === undefined) {
      return false;
    }
    return this.children[0].contains(value);
  } else if (value > this.value) {
    if (this.children[1] === undefined) {
      return false;
    }
    return this.children[1].contains(value);
  }
};

var a = new BinaryTree();
a.value = 10;
a.addChild(4);
a.addChild(11);
a.addChild(3);

BinaryTree.prototype.childrenArray = function() {
  var results = [];

  if (this.value) {
    results.push(this.value);
  }

  if (this.children[0].length === 0) {

    return results;
  }

  for (var i = 0; i < this.children[0].children.length; i++) {
    if (this.children[i].value) {
      results.push(this.children[i].value);
      return this.childrenArray();
    }
  }

};

a.childrenArray();

Answer 1

As @melpomene mentioned, you are invoking childArray but you didn't define it anywhere. I assume the line return childArray(); ended up there by mistake, you probably meant to recursively return childrenArray for the left child of root.

I would like to mention that your loop itself (without the recursive call):

for(var i = 0; i < this.children[0].children.length; i++) { // <-- you are iterating over the children of root's left child
     if(this.children[i].value) { // <-- but you are accessing root's current child
        results.push(this.children[i].value);
     }
}

is somewhat confusing. You are iterating over the children of root's left child children[0].children , but on each iteration you check if the root's children themselves children[i] have a value and you push that value. This is incorrect and will break if, for example, the root only has a left child that has both children ( i will be out of bounds).

---

Here's how you can approach this problem. The recursion to print the left tree in your case can be broken down into the following cases:

1. If the current node has no value, return an empty array
2. Else If the current node has no children, return an array only containing the current value
3. Else if the current node has a left child, run childrenArray on it recursively and add the results to the results array

Here's how that would look:

BinaryTree.prototype.childrenArray = function() {
  var results = [];

  // case 1:
  if (this.value === undefined) {
    return results;
  }

  // case 2:
  results.push(this.value);
  if (this.children.length === 0) {
    return results;
  }

  // case 3:
  var leftChild = this.children[0];
  if (leftChild) {
    results = results.concat(leftChild.childrenArray());
  }

  /* add code here for the right child to complete your function */

  return results;
};

Full Example:

 var BinaryTree = function(value) { this.value = value; this.children = []; }; BinaryTree.prototype.addChild = function (value) { if (value < this.value) { if (this.children[0] === undefined) { this.children[0] = new BinaryTree(value); } this.children[0].addChild(value); } else if (value > this.value) { if (this.children[1] === undefined) { this.children[1] = new BinaryTree(value); } this.children[1].addChild(value); } }; BinaryTree.prototype.childrenArray = function() { var results = []; // case 1: if (this.value === undefined) { return results; } // case 2: results.push(this.value); if (this.children.length === 0) { return results; } // case 3: var leftChild = this.children[0]; if (leftChild) { results = results.concat(leftChild.childrenArray()); } /* add code here for the right child to complete your function */ return results; }; var a = new BinaryTree(10); a.addChild(4); a.addChild(11); a.addChild(3); console.log(a.childrenArray()); // [10, 4, 3]

---

Last thing, based on your insertion logic (insert left if smaller, right if larger), you are building a Binary Search Tree not a Binary Tree . Binary tree's don't follow any particular insertion logic. Take a look at this question for clarification